Question

A pulley system lifts a load of 600N by an effort of 200N .If the resistance due to movable parts of a machineis 400N.FInd

1.M.A.

2.V.R.

3.Number of pulley

4.Efficiency

Answer 1

Given info : A pulley system lifts a load of 600N by an effort of 200N.

To find : if the resistance due to movable parts of mechanics is 400N, find

1. Machenical advantage
2. velocity ratio
3. number of pulleys
4. efficiency

solution : machenical advantage = load/effort

= 600N/200N = 3

velocity ratio = number of pulleys = machenical advantage = 3

actual machenical advantage = MA - (resistance/load) = 3 - 400/200 = 3 - 2 = 1

so efficiency = actual MA/VR × 100

= 1/3 × 100

= 33 1/3 %

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Answer 2

GIVEN :

Load(L) = 600 N

Resistance due to movable parts = 400 N

Effort (E) = 200 N

TO FIND :

• Mechanical advantage

• Velocity Ratio

• Number of pulleys

• Efficiency

$→Mechanical\: advantage = \frac{load}{effort} \\= \frac{600N}{200N} \\= 3$

$→Velocity\:Ratio(VR)=Number\: of \:pulleys\\=3$

$→Actual \:Mechanical\: Advantage = MA - \frac{Resistance}{Effort}\\= 3 - \frac{400}{200}\\= 3 - 2 \\= 1$

→Efficiency = Actual MA/VR × 100=1/3 × 100= 33.33 %

HERE IS YOUR ANSWER ☺️✌️