Question

Two snowcats in Antarctica are towing a housing unit north, as shown in Fig. 4–50. The sum of the forces${\overrightarrow{F}}_A$and${\overrightarrow{F}}_B$exerted on the unit by the horizontal cables is north, parallel to the line L, and$F_A=4500\mathrm{N}$Determine$F_B$and the magnitude of${\overrightarrow{F}}_{\mathbf{A}}+{\overrightarrow{F}}_{\mathbf{B}}$.

Short answer

The obtained values of forces are $F_B=6310.6\mathrm{N}$and $\Sigma F_y=8362.7\mathrm{N}$.

Step-by-step solution

Step 1. Calculate the forces in the x-direction

In this problem, consider the x-direction on the right side of the diagram and the y-direction as the forward direction for the snowcat’s motion.

There is no acceleration or net force in the x-direction because the housing unit travels on a straight line in the forward direction.

Given data:

The force on A is $F_A=4500\mathrm{N}$.

The relation to calculate the forces in the x-direction is given by:

$\begin{array}{c}\Sigma F_x=0\\ -F_A\sin 48°+F_B\sin 32°=0\\ F_B=\frac{F_A\sin 48°}{\sin 32°}\end{array}$

Here, $F_A$and $F_B$are the forces exerted on the unit by the cables.

On plugging the values in the above relation, you get:

$\begin{array}{l}{F}_B=\left[\frac{\left(4500\mathrm{N}\right)\sin 48°}{\sin 32°}\right]\\ F_B=6310.6\mathrm{N}\end{array}$

Step 2. Determine the magnitude of the sum of forces

The relation to calculate the forces in the y-direction is given by:

$\begin{array}{c}\Sigma F_y=0\\ \Sigma F_y=F_A\cos 48°+F_B\cos 32°\end{array}$

On plugging the values in the above relation, you get:

$\begin{array}{c}\Sigma F_y=\left(4500\mathrm{N}\right)\cos 48°+\left(6310.6\mathrm{N}\right)\cos 32°\\ \Sigma F_y=8362.7\mathrm{N}\end{array}$

Thus, the magnitude of the force on B is $F_B=6310.6\mathrm{N}$, and the magnitude of the sum of forces is $\Sigma F_y=8362.7\mathrm{N}$.