91Ó°ÊÓ

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

Calculate the net force applied in the horizontal direction as:

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

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:

$\begin{array}{c}{\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{,}\mathbf{y}}\mathbf{=}{\mathbf{F}}_{\mathbf{âŠ\yen }}\mathbf{−}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{−}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{=}\mathbf{0}\\ {\mathbf{F}}_{\mathbf{âŠ\yen }}\mathbf{=}\mathbf{2}\mathbf{T}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\end{array}$

Here ${\mathbf{F}}_{\mathbf{âŠ\yen }}$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,

$\begin{array}{c}{F}_{âŠ\yen }=2×12000\text{ N}×\sin 2^{°}\\ =24000\text{ N}×0.0349\\ =837.6\text{ N}\end{array}$

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

$T=\frac{F_{âŠ\yen }}{2\sin \mathrm{θ}}$

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

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

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