A screw jack with a pitch of 4\text{ mm} and a tommy bar of length 56\text{ cm} is used to… — JAMB UTME

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%Answer
B
400%
C
157%
D
2.5%

Answer

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

VR

and multiplying by

100\%

yields an efficiency of

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

Step-by-Step Solution

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.

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.

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

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%.

Key Concept

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

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