Suppose your car was mired deeply in the mud, and you wanted to use the method illustrated in Figure 4.37 to pull it out.

(a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

(b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?

• Force =12,000 N.
• The angle =2.00°.

Calculate the net force applied in the horizontal direction as:

$\begin{array}{c}{\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{,}\mathbf{x}}\mathbf{=}\mathbf{T}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{-}\mathbf{T}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\\ \mathbf{=}\mathbf{0}\end{array}$

Here, T is the tension in the rope, and θ is the angle made by the rope with the horizontal.

Write the expression for the net force applied in the vertical direction and equate it to zero as:

$\begin{array}{c}{\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{,}\mathbf{y}}\mathbf{=}{\mathbf{F}}_{\mathbf{\perp }}\mathbf{-}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{-}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\mathbf{0}\\ {\mathbf{F}}_{\mathbf{\perp }}\mathbf{=}\mathbf{2}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\end{array}$

Here ${\mathbf{F}}_{\mathbf{\perp }}$is the force exerted perpendicular to the center of the rope.

Substitute 12000 N for T and $2°$for θ in the above expression, and we get,

$\begin{array}{c}{F}_{\perp }=2\times 12000\text{\hspace{0.33em}N}\times \sin 2^{°}\\ =24000\text{\hspace{0.33em}N}\times 0.0349\\ =837.6\text{\hspace{0.33em}N}\end{array}$

Hence, the force exerted perpendicular to the rope is 837.6 N.

$T=\frac{F_{\perp }}{2\sin \theta }$

Substitute 837.6 N for $F_{\perp }$and $7°$for θ in the above expression, and we get,

$\begin{array}{c}T=\frac{837.6\text{\hspace{0.33em}N}}{2\sin 7^{°}}\\ =\frac{837.6\text{\hspace{0.33em}N}}{2\times 0.12187}\\ =3436.45\text{\hspace{0.33em}N}\end{array}$

Hence, the force exerted on the car is 3436.45 N.