A home mechanic wants to raise the 280-kg engine out of a car. The plan is to stretch a rope vertically from the engine to a branch of a tree 6.0 m above, and back to the bumper (Fig. 9–88). When the mechanic climbs up a stepladder and pulls horizontally on the rope at its midpoint, the engine rises out of the car. (a) How much force must the mechanic exert to hold the engine 0.50 m above its normal position? (b) What is the system’s mechanical advantage?

The mass of the car engine is\(m = 280\;{\rm{kg}}\).

The distance between the branch and the engine is \(d = 6.0\;{\rm{m}}\).

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it.

With the help of free body diagrams, one can accurately define the object of interest to which you are applying mechanical equations.

The free body shown below depicts the forces on the car engine and the forces at the ‘point of application of force’ on the rope the mechanic is pulling.

The forces acting on the system are the following:

• Weight of the car engine,\(W = mg\)vertically downwards
• Tension in the rope,\({F_{\rm{T}}}\)perpendicular to the floor
• Applied force (pulling), F

Initially, both the cables (part of the rope) are straight, and when the mechanic pulls the cable on the left toward the left direction, the engine is raised to 0.50 m.

This implies that there is a reduction of 0.50 m in the right cable and an extension. The extension is equally above and below the point of application of force.

Therefore, the hypotenuse of all the triangles shown in the FBD will have a length,

\(l = 3\;{\rm{m + }}\frac{{0.50\;{\rm{m}}}}{2} = 3.25\;{\rm{m}}\).

The length of the hypotenuse is the part of the rope from the tree branch to the mechanic.

The angle\(\theta \)can be determined as follows:

\(\begin{array}{c}\cos \theta = \frac{{3\;{\rm{m}}}}{{3.25\;{\rm{m}}}}\\\theta = {\cos ^{ - 1}}\left( {\frac{{3\;{\rm{m}}}}{{3.25\;{\rm{m}}}}} \right)\\ = {22.62^{\rm{o}}}\end{array}\)

At the point of application of force, i.e., along the x-direction, the net force will be given by the equation of equilibrium as follows,

\(\begin{array}{c}\sum {{F_{\rm{x}}}} = 0\\F - 2{F_{\rm{T}}}\sin \theta = 0\\F = 2{F_{\rm{T}}}\sin \theta \end{array}\) … (i)

The tension in the rope is equal to the weight of the car engine. Along the y-direction, the net force will be given by the following:

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = 0\\{F_{\rm{T}}} - mg = 0\\{F_{\rm{T}}} = mg\end{array}\) … (ii)

Substitute equation (ii) in (i) to determine the force F exerted by the mechanic to gold the engine.

\(\begin{array}{c}F = 2mg\sin \theta \\ = 2\left( {280\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\sin \left( {{{22.62}^{\rm{o}}}} \right)\\ = 2111\;{\rm{N}}\\ \approx 2100\;{\rm{N}}\end{array}\)

The mechanical advantage, in general, is used to calculate the increased force or the force gained by the mechanical system.

It is defined as the ratio of the externally applied force to the load (load force) and the force required to overcome the applied force.

Here, the mechanical advantage is given as follows:

\(\begin{array}{c}{\rm{MA}} = \frac{{{\rm{Load}}\;{\rm{force}}}}{{{\rm{Applied}}\;{\rm{force}}}}\\ = \frac{{mg}}{F}\\ = \frac{{\left( {280\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{2100\;{\rm{N}}}}\\ = 1.3\end{array}\)