Question

Two mechanics use pulleys to lift a 105 kg engine out of a car. The
first mechanic pulls with a force of 675 N (U21°R) and the second
mechanic pulls with a force of 712 N [U330L). The force of gravity
acting on the engine is 1029 N (D).
a) Draw the FBD for the engine.
b) Determine the net force acting on the engine.
c) Find the acceleration of the engine.
d) Determine how long it will take to lift the engine 1.25 m from rest.

Answer 1

(a) The free body diagram (FDB) consists of two upward forces and one downward force.

(b) The net force acting on the engine is 358 N.

(c) The acceleration of the engine is 3.41 m/s².

(d) The time taken to lift the engine from rest to the given height is 0.86 s.

Free body diagram (FBD) of the engine

The free body diagram of the engine is presented in the sketch below.

               

              F₁         F₂

               ↑        ↑

                ---------

                    ↓ W

Net force on the engine

The net force on the engine is calculated as follows;

$F_{net} = F_1 + F_2 - W\\\\F_{net} = 675 + 712 - 1029\\\\F_{net} = 358 \ N$

Acceleration of the engine

The acceleration of the engine is calculated as follows;

$a = \frac{F_{net}}{m} \\\\a = \frac{358}{105}\\\\ a = 3.41 \ m/s^2$

Time taken to lift the engine from rest

The time taken to lift the engine from rest is calculated using the following kinematic equation;

$s = v_0t + \frac{1}{2} at^2\\\\1.25 = 0 + \frac{1}{2} (3.41)t^2\\\\1.25 = 1.705t^2\\\\t^2 = \frac{1.25}{1.705} \\\\t^2 = 0.733\\\\t = \sqrt{0.733} \\\\t = 0.86 \ s$

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