How to Calculate Acceleration in a Pulley System for Lifting an Engine

[weedannycool]

Homework Statement

A car mechanic wants to lift the engine from a car using a rope and pulley
arrangement. He considers two options, shown in Figure Q1a and Figure Q1b. The car engine
has a mass, m, of 200kg, and the maximum force that the mechanic can exert on the rope, P, is
1500N. Use a free-body diagram approach to derive expressions for the acceleration
experienced by the car engine in each case. Use these expressions to calculate the vertical
acceleration of the engine if the mechanic exerts the maximum force on the rope. You may
neglect the mass of the pulley systems, and assume that they are frictionless.

Why can the man lift the engine in case b)(right of the figure). but not a)(left of figure)?

Homework Equations

$$\sum= Fy$$
$$\sum=Fx$$

The Attempt at a Solution

this is a dynamics question. i think wat i would do is find the sum of forces in the X and Y direction. however i am not sure. either the or it would be using circular accelertion.

Thanks

 

[PhanthomJay]

Attach figures, please.

 

[weedannycool]

soz for the rubbish drawing

 

[PhanthomJay]

That is mechanical advantage working for the mechanic. Draw a free body diagram of the engine in case a; then draw a free body diagram of the engine and pulley in case b. Use Newton's laws. Acceleration of the engine is not 'circular', it is 'linear' (it moves and accelerates straight up, if it is to move at all). Note that tensions in the rope on either side of an ideal pulley are the same.

 

[DeeNos]

for your question! Calculating acceleration in a pulley system involves understanding the forces acting on the system and using Newton's second law, F=ma. In this case, we have two different pulley systems, one in Figure Q1a and one in Figure Q1b. In both cases, the mechanic is exerting a force, P, on the rope, which is connected to the engine. However, in Figure Q1a, the rope is attached to the ceiling, while in Figure Q1b, the rope is attached to the ground.

To calculate the acceleration in each case, we need to consider the forces acting on the engine. In Figure Q1a, the engine is being pulled upward by the force of gravity, mg, and downward by the tension in the rope, T. The mechanic is exerting a force, P, on the rope in the upward direction. Using the free-body diagram approach, we can write the sum of forces in the Y direction as:

\sum F_y = T - mg - P = ma

Since we want to find the acceleration, we can rearrange this equation to solve for a:

a = (T - mg - P)/m

In Figure Q1b, the engine is also being pulled upward by the force of gravity, mg, but now the tension in the rope, T, is pulling in the opposite direction. The mechanic is still exerting a force, P, on the rope in the upward direction. Using the same approach as before, we can write the sum of forces in the Y direction as:

\sum F_y = -T - mg + P = ma

Solving for a, we get:

a = (-T - mg + P)/m

Now, we can calculate the acceleration for each case by plugging in the given values. Since we are assuming the pulley systems are frictionless, we can also assume that the tension, T, is equal in magnitude in both cases. Therefore, we can set the two equations equal to each other and solve for T:

(T - mg - P)/m = (-T - mg + P)/m

Solving for T, we get:

T = (P + mg)/2

Now, we can plug this value of T into our acceleration equations to find the acceleration for each case:

In Figure Q1a: a = ((P + mg)/2 - mg - P)/m = -P/2m