Mechanics Alıştırma Soruları — JAMB UTME — Sayfa 9

Soru 161Soru →

A U-tube open to the atmosphere at both ends contains water of density $1000\text{ kg m}^{-3}$ . An immiscible liquid is poured into one arm of the tube until it forms a column of height $15.0\text{ cm}$ . If the interface between the two liquids lies $12.0\text{ cm}$ below the surface of the water in the opposite arm, what is the density of the liquid in $\text{kg m}^{-3}$ ?

Cevabı ve açıklamayı göster

Cevap: 800

Cevap

The density of the immiscible liquid is

800\text{ kg m}^{-3}

.

At the level of the liquid interface, the hydrostatic pressure exerted by the

15.0\text{ cm}

liquid column must equal the hydrostatic pressure exerted by the

12.0\text{ cm}

water column above it. Equating

\rho_1 h_1 = \rho_2 h_2

yields

\rho_{\text{liquid}} = 1000 \times (12.0 / 15.0) = 800\text{ kg m}^{-3}

.

Adım Adım Çözüm

1

Apply the principal of equal pressure at the same horizontal level in a continuous fluid at rest.

Pressure due to liquid column equals pressure due to water column above the interface line.

Hydrostatic pressure at depth

h

is given by

P = \rho g h

, and points at equal depth in a connected liquid body share identical pressure.

2

Set up the density-height ratio relationship:

\rho_{\text{liquid}} \cdot h_{\text{liquid}} = \rho_{\text{water}} \cdot h_{\text{water}}

.

\rho_{\text{liquid}} = \rho_{\text{water}} \times \frac{h_{\text{water}}}{h_{\text{liquid}}}

.

Acceleration due to gravity

g

cancels out from both sides of the pressure balance equation.

3

Substitute

h_{\text{liquid}} = 15.0\text{ cm}

,

h_{\text{water}} = 12.0\text{ cm}

, and

\rho_{\text{water}} = 1000\text{ kg m}^{-3}

.

\rho_{\text{liquid}} = 1000 \times \frac{12.0}{15.0} = 800\text{ kg m}^{-3}

.

Heights can remain in centimeters since their unit ratio is dimensionless.

Anahtar Kavram

Hydrostatic Pressure Balance in U-Tube Manometers

Tahmini Süre:1m 30s

Soru 162Soru →

A uniform beam $PQ$ of length $4.0\text{ m}$ and mass $20\text{ kg}$ is supported horizontally on a pivot at end $P$ and by a vertical wire attached at end $Q$ . A load of mass $30\text{ kg}$ is placed on the beam at a distance of $1.0\text{ m}$ from $P$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the vertical wire attached at $Q$ in newtons?

Cevabı ve açıklamayı göster

Cevap: 175

Cevap

The tension in the vertical wire attached at end

Q

is

175\text{ N}

.

By applying the principle of moments about the pivot at

P

, the sum of downward clockwise moments produced by the

30\text{ kg}

load (

300\text{ N} \times 1.0\text{ m} = 300\text{ N}\cdot\text{m}

) and the beam's center of gravity (

200\text{ N} \times 2.0\text{ m} = 400\text{ N}\cdot\text{m}

) equals

700\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the tension force (

T \times 4.0\text{ m}

) gives

T = 175\text{ N}

.

Adım Adım Çözüm

1

Determine forces and their perpendicular distances from the pivot at

P

.

The load exerts a downward force of

300\text{ N}

at

1.0\text{ m}

from

P

. The uniform beam's weight of

200\text{ N}

acts at its midpoint (

2.0\text{ m}

from

P

). The vertical tension

T

acts upward at

Q

(

4.0\text{ m}

from

P

).

Before applying the principle of moments, all force magnitudes and their distance arms relative to the pivot point must be identified.

2

Equate total clockwise moments to total counterclockwise moments about

P

.

(300\text{ N} \times 1.0\text{ m}) + (200\text{ N} \times 2.0\text{ m}) = T \times 4.0\text{ m}

For rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of counterclockwise moments about that same pivot.

3

Calculate the value of the tension force

T

.

T = \frac{700}{4.0} = 175\text{ N}

Simplifying the moment equation gives the magnitude of the upward supporting force.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

Soru 163Soru →

A light rigid lever of length $2.5\text{ m}$ is pivoted horizontally at one end $O$ . A downward vertical weight of $60\text{ N}$ is hung from the lever at a distance of $1.5\text{ m}$ from $O$ . An upward force $F$ inclined at an angle of $30^\circ$ to the lever is applied at the free end. What is the magnitude of the force $F$ required to maintain horizontal equilibrium?

36\text{ N}

72\text{ N}

104\text{ N}

180\text{ N}

Cevabı ve açıklamayı göster

Cevap: $72\text{ N}$

Cevap

The magnitude of the force

F

required to maintain equilibrium is

72\text{ N}

.

According to the Principle of Moments, for a body in rotational equilibrium, the total clockwise moment about a pivot equals the total counterclockwise moment. The load creates a clockwise moment of

60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

. The force

F

applied at

30^\circ

has a perpendicular component of

F \sin 30^\circ = 0.5F

. The counterclockwise moment is

0.5F \times 2.5\text{ m} = 1.25F

. Setting

1.25F = 90\text{ N}\cdot\text{m}

gives

F = 72\text{ N}

.

Adım Adım Çözüm

1

Calculate the clockwise moment about the pivot

O

caused by the hanging load.

\text{Clockwise Moment} = 60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

The force of

60\text{ N}

acts perpendicularly at a distance of

1.5\text{ m}

from the pivot.

2

Express the counterclockwise moment about the pivot

O

exerted by the force

F

.

\text{Counterclockwise Moment} = F \sin(30^\circ) \times 2.5\text{ m} = 1.25 F\text{ N}\cdot\text{m}

Only the perpendicular component of the force,

F \sin(30^\circ)

, contributes to the moment about pivot

O

.

3

Equate the clockwise and counterclockwise moments according to the Principle of Moments.

1.25 F = 90 \implies F = \frac{90}{1.25} = 72\text{ N}

For rotational equilibrium, the sum of clockwise moments must equal the sum of counterclockwise moments.

Anahtar Kavram

Principle of Moments and Perpendicular Force Components

Soru 164Soru →

Planet $P$ has a mean uniform density that is twice that of Planet $Q$ , and a radius that is half that of Planet $Q$ . What is the ratio of the escape velocity at the surface of Planet $P$ to the escape velocity at the surface of Planet $Q$ ?

\frac{1}{\sqrt{2}}

\frac{1}{2}

1

\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{\sqrt{2}}$

Cevap

\frac{1}{\sqrt{2}}

The correct answer is derived by substituting the planet's mass in terms of radius and density (

M = \frac{4}{3}\pi R^3 \rho

) into the escape velocity equation

v_e = \sqrt{\frac{2GM}{R}}

. Simplifying gives

v_e = R\sqrt{\frac{8}{3}\pi G \rho}

, which means escape velocity is proportional to

R\sqrt{\rho}

. Halving the radius and doubling the density yields a factor of

\frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

Adım Adım Çözüm

1

Express the mass of a uniform spherical planet in terms of its radius

R

and density

\rho

.

M = \frac{4}{3}\pi R^3 \rho

Mass equals volume multiplied by mean density.

2

Substitute mass into the escape velocity formula

v_e = \sqrt{\frac{2GM}{R}}

.

v_e = \sqrt{\frac{2G \left(\frac{4}{3}\pi R^3 \rho\right)}{R}} = R\sqrt{\frac{8}{3}\pi G \rho}

This establishes the proportionality

v_e \propto R\sqrt{\rho}

.

3

Set up the ratio of escape velocity for Planet

P

to Planet

Q

using

R_P = \frac{1}{2}R_Q

and

\rho_P = 2\rho_Q

.

\frac{v_{e,P}}{v_{e,Q}} = \frac{R_P\sqrt{\rho_P}}{R_Q\sqrt{\rho_Q}} = \frac{\left(\frac{1}{2}R_Q\right)\sqrt{2\rho_Q}}{R_Q\sqrt{\rho_Q}} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}}

Evaluating the ratio of proportional quantities yields the final scale factor.

Anahtar Kavram

Dependence of Escape Velocity on Planet Density and Radius

Tahmini Süre:2m 0s

Soru 165Soru →

A satellite revolves around a planet in a circular orbit of radius $1.0 \times 10^4 \text{ km}$ with an orbital period of $12 \text{ hours}$ . Calculate the orbital period, in hours, of a second satellite orbiting the same planet in a circular path of radius $4.0 \times 10^4 \text{ km}$ .

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

96 hours

According to Kepler's Third Law (

T^2 \propto r^3

), the orbital period

T

scales with radius

r

as

T \propto r^{3/2}

. Increasing the orbital radius by a factor of 4 increases the period by a factor of

4^{3/2} = 8

. Multiplying the original period of 12 hours by 8 yields 96 hours.

Adım Adım Çözüm

1

Set up Kepler's Third Law equation relating orbital period and orbital radius.

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

Kepler's Third Law states that the square of the orbital period of a body in circular orbit is directly proportional to the cube of the radius of its orbit.

2

Substitute the given orbital radii and evaluate the scaling factor.

\frac{r_2}{r_1} = \frac{4.0 \times 10^4 \text{ km}}{1.0 \times 10^4 \text{ km}} = 4

Simplifying the ratio of the two orbital radii gives a factor of 4 increase in radius.

3

Calculate the period multiplier by taking the ratio to the power of 3/2.

4^{3/2} = (\sqrt{4})^3 = 2^3 = 8

Taking

T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2}

shows the period scales by a factor of 8.

4

Multiply the initial orbital period by the scaling factor to find the final answer.

T_2 = 12 \text{ hours} \times 8 = 96 \text{ hours}

Multiplying the baseline period of 12 hours by 8 yields the new orbital period.

Anahtar Kavram

Kepler's Third Law of Planetary Motion

Tahmini Süre:1m 30s

Soru 166Soru →

A uniform horizontal beam $AB$ of length $4.0\text{ m}$ and mass $10\text{ kg}$ is hinged smoothly to a vertical wall at end $A$ . It is held horizontally in static equilibrium by a light cable attached to end $B$ and anchored to the wall above $A$ , making an angle of $30^\circ$ with the beam. A mass of $5\text{ kg}$ is suspended from the beam at a distance of $3.0\text{ m}$ from hinge $A$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the cable?

175\text{ N}

87.5\text{ N}

275\text{ N}

17.5\text{ N}

Cevabı ve açıklamayı göster

Cevap: $175\text{ N}$

Cevap

The tension in the cable is

175\text{ N}

.

Applying the principle of moments about the hinge at end A, the clockwise moments due to the beam's center of mass (

100\text{ N}

at

2.0\text{ m}

) and the suspended load (

50\text{ N}

at

3.0\text{ m}

) are balanced by the counterclockwise moment of the cable tension (

T \sin 30^\circ

at

4.0\text{ m}

). Solving

(100 \times 2.0) + (50 \times 3.0) = 2.0 T

yields

T = 175\text{ N}

.

Adım Adım Çözüm

1

Calculate the downward gravitational forces (weights) acting on the system.

Weight of beam

W_{\text{beam}} = m_{\text{beam}} g = 10\text{ kg} \times 10\text{ m/s}^2 = 100\text{ N}

acting at

2.0\text{ m}

from

A

. Weight of load

W_{\text{load}} = m_{\text{load}} g = 5\text{ kg} \times 10\text{ m/s}^2 = 50\text{ N}

acting at

3.0\text{ m}

from

A

.

Forces causing clockwise moments must be expressed in force units (newtons) and located at their respective lines of action.

2

Formulate the equilibrium condition using the Principle of Moments about hinge

A

.

\sum \tau_A = 0 \implies (W_{\text{beam}} \times 2.0\text{ m}) + (W_{\text{load}} \times 3.0\text{ m}) = T \sin(30^\circ) \times 4.0\text{ m}

The hinge

A

eliminates reaction forces at the hinge from the moment equation.

3

Substitute numerical values and solve for tension

T

.

(100 \times 2.0) + (50 \times 3.0) = T \times 0.5 \times 4.0 \implies 200 + 150 = 2.0 T \implies 350 = 2.0 T \implies T = 175\text{ N}$.

Perpendicular distance from

A

to line of action of tension is

4.0 \sin 30^\circ = 2.0\text{ m}

.

Anahtar Kavram

Equilibrium of rigid bodies and Principle of Moments under non-perpendicular forces

Tahmini Süre:2m 0s

Soru 167Soru →

A satellite of mass $500\text{ kg}$ orbits a spherical planet of radius $R = 6.0 \times 10^6\text{ m}$ with surface gravitational acceleration $g = 10\text{ m/s}^2$ . The satellite is transferred from an initial circular orbit of radius $2R$ to a higher circular orbit of radius $3R$ . What is the minimum energy required, in megajoules ( $\text{MJ}$ ), to perform this transfer?

Cevabı ve açıklamayı göster

Cevap: 2500

Cevap

The minimum energy required to perform the orbital transfer is

2500\text{ MJ}

.

The minimum energy needed to move a satellite between circular orbits is equal to the change in its total mechanical energy (

E = -\frac{GMm}{2r}

). Expressing

GM

as

gR^2

, the energy difference between radii

2R

and

3R

simplifies to

\Delta E = \frac{gRm}{12}

, which evaluates to

2500\text{ MJ}

.

Adım Adım Çözüm

1

Relate surface acceleration due to gravity to planet mass and radius.

GM = gR^2

At the planet's surface (

r = R

), gravitational acceleration is

g = \frac{GM}{R^2}

.

2

Formulate the total mechanical energy equation for a circular orbit.

E = -\frac{GMm}{2r} = -\frac{gR^2 m}{2r}

Total energy is kinetic energy

\frac{GMm}{2r}

plus gravitational potential energy

-\frac{GMm}{r}

.

3

Calculate initial and final total energies.

E_1 = -\frac{gRm}{4}

and

E_2 = -\frac{gRm}{6}

Substitute the orbit radii

r_1 = 2R

and

r_2 = 3R

into the total energy equation.

4

Determine the net work required for the transfer.

\Delta E = E_2 - E_1 = \frac{gRm}{12}

The energy required equals the difference in total mechanical energy between the final and initial orbits.

5

Substitute given numerical values and convert joules to megajoules.

\Delta E = \frac{10 \times (6.0 \times 10^6) \times 500}{12} = 2.5 \times 10^9\text{ J} = 2500\text{ MJ}

Dividing

2.5 \times 10^9\text{ J}

by

10^6

converts the value to megajoules.

Anahtar Kavram

Total Mechanical Energy of a Satellite in Circular Orbit and Orbital Transfer Energy

Tahmini Süre:3m 0s

Soru 168Soru →

An astronaut has a weight of $720\text{ N}$ on the surface of the Earth. Calculate the weight of the astronaut, in Newtons ( $\text{N}$ ), at an altitude equal to twice the radius of the Earth ( $h = 2R$ ).

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The weight of the astronaut at an altitude of

2R

is

80\text{ N}

.

At an altitude of

2R

, the total distance from the center of the Earth is

r = R + 2R = 3R

. Because gravitational force follows the inverse-square law (

W \propto 1/r^2

), tripling the distance reduces the gravitational force and weight by a factor of

3^2 = 9

. Dividing the surface weight of

720\text{ N}

by

9

yields

80\text{ N}

.

Adım Adım Çözüm

1

Determine the total distance from the center of the Earth

r = R + 2R = 3R

Gravitational force depends on the distance measured from the center of mass of the Earth, which is the sum of Earth's radius

R

and altitude

h

.

2

Apply the inverse-square law of gravitation to find field strength at altitude

g' = \frac{g}{3^2} = \frac{g}{9}

Acceleration due to gravity is inversely proportional to the square of the distance from the planet's center (

g \propto \frac{1}{r^2}

).

3

Calculate the astronaut's weight at altitude

W' = \frac{720}{9} = 80\text{ N}

Weight is directly proportional to gravitational field strength (

W = mg

).

Anahtar Kavram

Variation of Acceleration due to Gravity with Altitude (Inverse Square Law)

Soru 169Soru →

Two point masses of $4.0\text{ kg}$ and $9.0\text{ kg}$ are separated by a distance of $5.0\text{ m}$ in free space. At what distance from the $4.0\text{ kg}$ mass along the line joining them is the net gravitational field strength equal to zero?

2.0\text{ m}

3.0\text{ m}

2.5\text{ m}

1.0\text{ m}

Cevabı ve açıklamayı göster

Cevap: $2.0\text{ m}$

Cevap

The distance from the

4.0\text{ kg}

mass where the net gravitational field strength is zero is

2.0\text{ m}

.

At the point where the net gravitational field strength is zero, the gravitational field intensity produced by the

4.0\text{ kg}

mass must equal the intensity produced by the

9.0\text{ kg}

mass in magnitude. Equating

\frac{G (4.0)}{x^2} = \frac{G (9.0)}{(5.0 - x)^2}

and taking square roots yields

\frac{2}{x} = \frac{3}{5 - x}

. Cross-multiplying gives

10 - 2x = 3x

, which yields

x = 2.0\text{ m}

from the

4.0\text{ kg}

mass.

Adım Adım Çözüm

1

Set up the condition for zero net gravitational field strength.

E_1 = E_2 \implies \frac{G m_1}{x^2} = \frac{G m_2}{(d - x)^2}

At the neutral point, the opposing gravitational field vectors due to both masses are equal in magnitude.

2

Substitute given values

m_1 = 4.0\text{ kg}

,

m_2 = 9.0\text{ kg}

, and total distance

d = 5.0\text{ m}

.

\frac{4.0}{x^2} = \frac{9.0}{(5.0 - x)^2}

Simplifying by canceling

G

from both sides of the equation.

3

Take the square root of both sides and solve for

x

.

\frac{2.0}{x} = \frac{3.0}{5.0 - x} \implies 2.0(5.0 - x) = 3.0x \implies 10.0 - 2.0x = 3.0x \implies 5.0x = 10.0 \implies x = 2.0\text{ m}

Taking the square root simplifies the quadratic relationship into a linear ratio.

Anahtar Kavram

Gravitational Field Strength Neutral Point

Tahmini Süre:1m 0s

Soru 170Soru →

A toy car of mass $0.5\text{ kg}$ moves along a smooth horizontal surface at a constant speed of $4\text{ m s}^{-1}$ . What is the kinetic energy of the toy car?

1\text{ J}

4\text{ J}

8\text{ J}

16\text{ J}

Cevabı ve açıklamayı göster

Cevap: $4\text{ J}$

Cevap

The kinetic energy of the toy car is

4\text{ J}

.

Substituting the given values into the formula

E_k = \frac{1}{2} m v^2

gives

E_k = \frac{1}{2} \times 0.5 \times 4^2 = 4\text{ J}

.

Adım Adım Çözüm

1

Identify the given physical quantities

Mass

m = 0.5\text{ kg}

and speed

v = 4\text{ m s}^{-1}

.

These are the mandatory inputs needed to determine kinetic energy.

2

Apply the kinetic energy formula

E_k = \frac{1}{2} m v^2

Kinetic energy of a translational body is defined as half the product of its mass and the square of its speed.

3

Substitute the values and compute the result

E_k = \frac{1}{2} \times 0.5\text{ kg} \times (4\text{ m s}^{-1})^2 = 0.25 \times 16 = 4\text{ J}

.

Evaluating the mathematical operation yields the energy in Joules.

Anahtar Kavram

Kinetic Energy

Tahmini Süre:45s

Soru 171Soru →

A screw jack with a pitch of $4\text{ mm}$ and a tommy bar of length $56\text{ cm}$ is used to raise a heavy load of mass $1100\text{ kg}$ . If an effort force of $50\text{ N}$ is applied at the outer end of the tommy bar, what is the efficiency of the screw jack? (Take $\pi = \frac{22}{7}$ and $g = 10\text{ m/s}^2$ )

25%

400%

157%

2.5%

Cevabı ve açıklamayı göster

Cevap: 25%

Cevap

The efficiency of the screw jack is 25%

The mechanical advantage is calculated as

MA = \frac{11,000\text{ N}}{50\text{ N}} = 220

. The velocity ratio for a screw jack is

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = 880

. Dividing

MA

by

VR

and multiplying by

100\%

yields an efficiency of

\frac{220}{880} \times 100\% = 25\%

.

Adım Adım Çözüm

1

Calculate the load force (weight) from the given mass.

W = m \times g = 1100\text{ kg} \times 10\text{ m/s}^2 = 11,000\text{ N}

Mass must be converted to weight force in newtons to compute mechanical advantage.

2

Calculate the Mechanical Advantage (MA).

MA = \frac{\text{Load}}{\text{Effort}} = \frac{11,000\text{ N}}{50\text{ N}} = 220

Mechanical advantage is the ratio of output load force to input effort force.

3

Calculate the Velocity Ratio (VR) of the screw jack.

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = \frac{3.52\text{ m}}{0.004\text{ m}} = 880

Velocity ratio is the distance moved by the effort in one full revolution (

2\pi R

) divided by the distance moved by the load in one revolution (pitch

p

).

4

Calculate the efficiency of the machine.

\text{Efficiency} = \frac{MA}{VR} \times 100\% = \frac{220}{880} \times 100\% = 25\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Anahtar Kavram

Mechanical Advantage, Velocity Ratio, and Efficiency of a Screw Jack

Soru 172Soru →

An object undergoing simple harmonic motion completes $20$ complete oscillations in a time duration of $10.0\text{ s}$ . What is the period of oscillation of the object in seconds?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The period of oscillation is

0.5\text{ s}

.

The period of simple harmonic motion is the time taken to complete one single oscillation. Dividing the total time (

10.0\text{ s}

) by the number of oscillations (

20

) yields

0.5\text{ s}

.

Adım Adım Çözüm

1

Apply the definition of oscillation period

T = \frac{t}{N}

Period

T

measures the time required for a single complete cycle.

2

Calculate the numerical value for period

T = \frac{10.0\text{ s}}{20} = 0.5\text{ s}

Dividing total elapsed time by total completed oscillations gives time per oscillation.

Anahtar Kavram

Period of Simple Harmonic Motion

Soru 173Soru →

A conveyor system pulls a $40\text{ kg}$ crate at a constant speed of $3\text{ m s}^{-1}$ up a rough inclined ramp. The ramp rises $3\text{ m}$ for every $5\text{ m}$ measured along its slope (giving $\sin\theta = 0.6$ and $\cos\theta = 0.8$ ). If the coefficient of kinetic friction between the crate and the ramp is $0.25$ and $g = 10\text{ m s}^{-2}$ , what is the power output of the conveyor system in watts?

Cevabı ve açıklamayı göster

Cevap: 960

Cevap

The power output required by the conveyor system is

960\text{ W}

.

To pull the crate up the ramp at constant speed, the conveyor force must overcome both the parallel gravitational component (

240\text{ N}

) and friction (

80\text{ N}

), making the total force

320\text{ N}

. Multiplying this force by the constant speed of

3\text{ m s}^{-1}

gives a total power output of

960\text{ W}

.

Adım Adım Çözüm

1

Calculate the component of weight parallel to the inclined plane

F_g = mg \sin\theta = 40 \times 10 \times 0.6 = 240\text{ N}

Gravity pulls the object back down along the slope with force

mg \sin\theta

.

2

Calculate the normal reaction force perpendicular to the plane

N = mg \cos\theta = 40 \times 10 \times 0.8 = 320\text{ N}

The normal force balances the perpendicular weight component.

3

Determine the magnitude of kinetic friction force

f_k = \mu N = 0.25 \times 320 = 80\text{ N}

Friction opposes motion up the slope and depends on the normal force.

4

Calculate the total pulling force needed for zero net acceleration

F = F_g + f_k = 240 + 80 = 320\text{ N}

At constant velocity, net force along the incline must equal zero, so

F = mg\sin\theta + f_k

.

5

Calculate the power output of the conveyor

P = F \times v = 320\text{ N} \times 3\text{ m s}^{-1} = 960\text{ W}

Power developed by a constant force moving an object at velocity

v

is given by

P = Fv

.

Anahtar Kavram

Work done against gravity and friction, and rate of doing work (Power

P = Fv

)

Soru 174Soru →

A wheel and axle system consists of a wheel with a radius of $25\text{ cm}$ attached to an axle with a radius of $5\text{ cm}$ . What is the velocity ratio of this machine?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The velocity ratio of the machine is 5.

The velocity ratio (VR) of a wheel and axle is the ratio of the radius of the wheel (

R

) to the radius of the axle (

r

). Calculating

\frac{25\text{ cm}}{5\text{ cm}}

gives a velocity ratio of

5

.

Adım Adım Çözüm

1

Identify the formula for the velocity ratio of a wheel and axle system.

\text{VR} = \frac{R}{r}

Velocity ratio is defined as the distance moved by the effort (proportional to wheel radius) divided by the distance moved by the load (proportional to axle radius).

2

Substitute the given values into the formula.

\text{VR} = \frac{25\text{ cm}}{5\text{ cm}}

The radius of the wheel

R = 25\text{ cm}

and the radius of the axle

r = 5\text{ cm}

.

3

Calculate the final ratio.

\text{VR} = 5

Dividing 25 by 5 yields 5. The ratio is dimensionless because the units of centimeters cancel out.

Anahtar Kavram

Velocity Ratio of a Wheel and Axle

Soru 175Soru →

A uniform horizontal wooden rod $XY$ of length $3.0\text{ m}$ and weight $50\text{ N}$ rests horizontally on two smooth supports located at $X$ (the left end) and at a point $Z$ which is $0.6\text{ m}$ from end $Y$ . A block of weight $120\text{ N}$ is placed on the rod at a distance of $0.9\text{ m}$ from end $X$ . What is the magnitude of the upward reaction force, in newtons, at support $Z$ ?

Cevabı ve açıklamayı göster

Cevap: 76.25

Cevap

The magnitude of the upward reaction force at support

Z

is

76.25\text{ N}

.

Taking moments about support

X

, the total clockwise moment is the sum of the moment due to the load (

120\text{ N} \times 0.9\text{ m} = 108\text{ N}\cdot\text{m}

) and the weight of the rod (

50\text{ N} \times 1.5\text{ m} = 75\text{ N}\cdot\text{m}

), giving

183\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the reaction force at

Z

(

R_Z \times 2.4\text{ m}

) yields

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

Adım Adım Çözüm

1

Identify the perpendicular distance of each force and support from pivot point

X

.

Center of gravity position

x_{cg} = 1.5\text{ m}

, load position

x_{L} = 0.9\text{ m}

, and support

Z

position

x_{Z} = 3.0 - 0.6 = 2.4\text{ m}

.

Taking moments about

X

requires knowing the exact moment arm for each force from

X

.

2

Set up the equation for rotational equilibrium about point

X

.

Total clockwise moment =

(120 \times 0.9) + (50 \times 1.5) = 183\text{ N}\cdot\text{m}

; Total counterclockwise moment =

R_Z \times 2.4

.

Choosing pivot

X

eliminates the unknown reaction force

R_X

because its distance from

X

is zero.

3

Equate clockwise moments to counterclockwise moments and solve for

R_Z

.

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

For a body in rotational equilibrium, the algebraic sum of moments about any point must equal zero.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

Soru 176Soru →

A wheel and axle machine has a wheel of radius $25\text{ cm}$ and an axle of radius $5\text{ cm}$ . If an effort force of $100\text{ N}$ is applied to lift a load of $400\text{ N}$ , what is the efficiency of the machine?

80\%

125\%

20\%

5\%

Cevabı ve açıklamayı göster

Cevap: $80\%$

Cevap

The efficiency of the wheel and axle machine is

80\%

.

First, calculate the velocity ratio as the ratio of wheel radius to axle radius:

\text{VR} = \frac{25}{5} = 5

. Next, calculate mechanical advantage as load over effort:

\text{MA} = \frac{400}{100} = 4

. Finally, calculate efficiency by dividing mechanical advantage by velocity ratio:

\text{Efficiency} = \frac{4}{5} \times 100\% = 80\%

.

Adım Adım Çözüm

1

Calculate the Velocity Ratio (VR) of the wheel and axle system

\text{VR} = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{25\text{ cm}}{5\text{ cm}} = 5

For a wheel and axle, the velocity ratio is the ratio of the radius of the wheel to the radius of the axle.

2

Calculate the Mechanical Advantage (MA) of the machine

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical advantage measures the force magnification of a machine, defined as the ratio of load raised to effort applied.

3

Calculate the Efficiency of the machine

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{4}{5} \times 100\% = 80\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Anahtar Kavram

Efficiency of Simple Machines

Soru 177Soru →

A hydraulic press has a small piston with a diameter of $4\text{ cm}$ and a large piston with a diameter of $20\text{ cm}$ . If an effort force of $80\text{ N}$ applied to the small piston raises a load of $1500\text{ N}$ placed on the large piston, what is the efficiency of the machine?

133.3\%

75.0\%

375.0\%

26.7\%

Cevabı ve açıklamayı göster

Cevap: $75.0\%$

Cevap

The efficiency of the hydraulic press is

75.0\%

.

The mechanical advantage is

MA = \frac{1500}{80} = 18.75

. The velocity ratio for pistons of diameters

20\text{ cm}

and

4\text{ cm}

is

VR = \left(\frac{20}{4}\right)^2 = 25

. Efficiency is

\frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

.

Adım Adım Çözüm

1

Calculate Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{1500\text{ N}}{80\text{ N}} = 18.75

Mechanical advantage is defined as the ratio of load force to effort force.

2

Calculate Velocity Ratio (VR) for the hydraulic press

VR = \frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{20\text{ cm}}{4\text{ cm}}\right)^2 = 5^2 = 25

The velocity ratio of a hydraulic press equals the ratio of the cross-sectional areas of the pistons, which simplifies to the square of the ratio of their diameters.

3

Calculate Efficiency (\eta)

\eta = \frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Anahtar Kavram

Hydraulic Press Efficiency and Velocity Ratio

Soru 178Soru →

A light, rigid horizontal bar $AB$ of length $1.5\text{ m}$ is smoothly pivoted at end $A$ . A vertical downward load of $40\text{ N}$ is hung from end $B$ . The bar is kept in horizontal equilibrium by a light string attached at point $C$ , located $1.0\text{ m}$ from $A$ . The string exerts a tension force $T$ pulling upwards at an angle of $30^\circ$ relative to the horizontal bar. What is the magnitude of the tension $T$ in newtons?

Cevabı ve açıklamayı göster

Cevap: 120

Cevap

The magnitude of the tension

T

in the string is

120\text{ N}

.

For the bar to maintain rotational equilibrium, the clockwise moment about pivot

A

must equal the counterclockwise moment about

A

. The

40\text{ N}

load exerts a clockwise moment of

40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

. The string tension

T

exerts a counterclockwise moment given by its vertical component multiplied by the distance from the pivot:

(T \sin 30^\circ) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

. Equating the two moments gives

0.5T = 60\text{ N}\cdot\text{m}

, yielding

T = 120\text{ N}

.

Adım Adım Çözüm

1

Calculate the clockwise moment about the pivot at end

A

τ_{\text{clockwise}} = 40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

The weight at

B

acts vertically downward at a perpendicular distance of

1.5\text{ m}

from pivot

A

.

2

Determine the perpendicular component of tension

T

relative to the bar

F_{\perp} = T \sin 30^\circ = 0.5T

Only the component of force perpendicular to the bar produces a moment about the pivot.

3

Set up the counterclockwise moment about pivot

A

τ_{\text{counterclockwise}} = (0.5T) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

The string is attached at point

C

, which is

1.0\text{ m}

away from pivot

A

.

4

Apply the principle of moments for rotational equilibrium and solve for

T

0.5T = 60 \implies T = 120\text{ N}

For rotational equilibrium, total clockwise moments must equal total counterclockwise moments about any pivot.

Anahtar Kavram

Principle of moments and rotational equilibrium for forces acting at non-perpendicular angles.

Tahmini Süre:1m 30s

Soru 179Soru →

A car of mass $1200\text{ kg}$ ascends a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$ , at a steady speed of $15\text{ m s}^{-1}$ . If the total frictional resistance to motion is $400\text{ N}$ , what is the useful mechanical power output of the engine in kilowatts? (Take $g = 10\text{ m s}^{-2}$ )

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

24 kW

To maintain a constant ascending speed, the engine must supply a force equal to the sum of the component of weight parallel to the incline (

mg\sin\theta = 1200\text{ N}

) and the frictional force (

400\text{ N}

), resulting in a total force of

1600\text{ N}

. Multiplying this total force by the constant speed of

15\text{ m s}^{-1}

yields a power output of

24000\text{ W}

, which corresponds to

24\text{ kW}

.

Adım Adım Çözüm

1

Calculate the gravitational force component acting down the slope

1200 N

The component of the car's weight parallel to the incline opposes upward motion:

F_g = m g \sin\theta = 1200 \text{ kg} \times 10 \text{ m s}^{-2} \times 0.1 = 1200 \text{ N}

.

2

Determine the total tractive force required from the car engine

1600 N

Because the velocity is constant, the net force is zero; hence, the engine force must balance both the gravitational slope component and the frictional resistance:

F_{\text{engine}} = F_g + F_{\text{friction}} = 1200 \text{ N} + 400 \text{ N} = 1600 \text{ N}

.

3

Calculate the mechanical power delivered by the engine

24 kW

Power is the product of tractive force and constant speed:

P = F_{\text{engine}} \times v = 1600 \text{ N} \times 15 \text{ m s}^{-1} = 24000 \text{ W} = 24 \text{ kW}

.

Anahtar Kavram

Power required to maintain motion against opposing forces on an inclined plane

Soru 180Soru →

A porter carries a suitcase of mass $5\text{ kg}$ along a horizontal platform for a distance of $10\text{ m}$ at a constant speed, and then lifts it vertically upward through a height of $2\text{ m}$ onto a shelf. Taking $g = 10\text{ m s}^{-2}$ , what is the total work done by the porter on the suitcase?

100\text{ J}

600\text{ J}

500\text{ J}

10\text{ J}

Cevabı ve açıklamayı göster

Cevap: $100\text{ J}$

Cevap

100\text{ J}

During horizontal motion at constant speed, the upward force exerted by the porter is perpendicular to the horizontal displacement, so no work is performed horizontally (

W_{\text{horizontal}} = F s \cos 90^\circ = 0\text{ J}

). During vertical lifting, the force acts in the direction of displacement, yielding

W_{\text{vertical}} = mgh = 5\text{ kg} \times 10\text{ m s}^{-2} \times 2\text{ m} = 100\text{ J}

. The total work done is therefore

100\text{ J}

.

Adım Adım Çözüm

1

Calculate the work done during horizontal motion

W_{\text{horizontal}} = 0\text{ J}

The supporting force acts vertically upward at an angle of

90^\circ

to the horizontal displacement, giving

W = F s \cos 90^\circ = 0\text{ J}

.

2

Calculate the work done in lifting the suitcase vertically

W_{\text{vertical}} = mgh = 5 \times 10 \times 2 = 100\text{ J}

Work done against gravity equals the increase in gravitational potential energy.

3

Sum the work done in both stages

W_{\text{total}} = 0 + 100 = 100\text{ J}

Total work done is the scalar sum of work performed along each segment of motion.

Anahtar Kavram

Work done by a constant force depends on the direction of displacement (

W = F s \cos \theta

). Perpendicular forces do no work.

Mechanics

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