A screw jack with a thread pitch of 0.5\text{ cm} is operated using a Tommy bar of length … — JAMB UTME

A screw jack with a thread pitch of $0.5\text{ cm}$ is operated using a Tommy bar of length $35\text{ cm}$ . If the machine has an efficiency of $40\%$ , what is the minimum effort force required to raise a load of mass $880\text{ kg}$ ? (Take $g = 10\text{ m/s}^2$ and $\pi = \frac{22}{7}$ )

$50\text{ N}$ Answer
B
$8\text{ N}$
C
$20\text{ N}$
D
$100\text{ N}$

Answer

The minimum effort force required is

50\text{ N}

The effort distance per turn is the circumference

2\pi r = 2 \times \frac{22}{7} \times 35\text{ cm} = 220\text{ cm}

. Dividing by the pitch (

0.5\text{ cm}

) yields a Velocity Ratio of

440

. Applying the

40\%

efficiency gives a Mechanical Advantage of

\text{MA} = 0.40 \times 440 = 176

. Finally, dividing the load force (

8800\text{ N}

) by

\text{MA}

gives an effort force of

50\text{ N}

Step-by-Step Solution

Calculate the total load force in Newtons

L = m \cdot g = 880\text{ kg} \times 10\text{ m/s}^2 = 8800\text{ N}

Effort overcomes weight force, which is mass multiplied by gravitational acceleration.

Calculate the Velocity Ratio (VR) of the screw jack

\text{VR} = \frac{2 \pi r}{p} = \frac{2 \times \frac{22}{7} \times 35\text{ cm}}{0.5\text{ cm}} = \frac{220\text{ cm}}{0.5\text{ cm}} = 440

The effort travels around the circumference of a circle of radius

r

, while the load moves vertically by one pitch length

p

per revolution.

Determine the Mechanical Advantage (MA) from efficiency

\text{MA} = \eta \times \text{VR} = 0.40 \times 440 = 176

Efficiency is defined as

\eta = \frac{\text{MA}}{\text{VR}}

, so

\text{MA} = \eta \cdot \text{VR}

Calculate the effort force E

E = \frac{L}{\text{MA}} = \frac{8800\text{ N}}{176} = 50\text{ N}

Mechanical advantage is the ratio of load to effort force

(\text{MA} = \frac{L}{E})

Key Concept

Relationship between Velocity Ratio, Mechanical Advantage, Efficiency, and Pitch in a Screw Jack

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