Question

Two ropes are attached to a heavy box to pull it along the floor. One rope applies a force of 333 newtons (N) in a direction due west; the other applies a force of 590 N in a direction due south. As we will see later in the text, force is a vector quantity. (a) How much force should be applied by a single rope, and (b) in what direction (as a positive angle relative to due west), if it is to accomplish the same effect as the two forces added together?

Answer 1

Answer:

(a) 677.49 N

(b) $60.56^\circ$

Explanation:

Given:

• $\vec{F}_1$ = force in the first rope = 333 N west = $-333\ N\ \hat{i}$

• $\vec{F}_2$ = force in the second rope = 590 N south = $-590\ N\ \hat{j}$

Assume:

• $\vec{F}_{net}$ = force in a single rope
• $\theta$ = angle with the west direction

We know force as a vector quantity. The two forces acting on the heavy box will have a resultant force whose magnitude and direction will be equivalent to the force required in a single rope that would produce the same effect on the box.

Let us first try to find out the resultant force.

Since the resultant of a force is calculated by the vector addition of all the force vectors.

$\therefore \vec{F}_{net} = \vbec{F}_1+\vec{F}_2\\\Rightarrow \vec{F}_{net} =(-333\ N\ \hat{i})+(-590\ N\ \hat{j})\\\Rightarrow \vec{F}_{net} =-333\ N\ \hat{i}-590\ N\ \hat{j}$

Part (a):

$\textrm{The magnitude of the force in that single rope}=\sqrt{(-333)^2+(-590)^2}\\\Rightarrow F_{net}= 677.49\ N$

Hence, a force of 677.49 N should be applied by a single rope to do the same effect.

Part (b):

Since the resultant force vector is has its coordinates in the third quadrant of the cartesian vector plane. So, the vector will absolutely make a positive angle with the west direction which is given by:

$\theta = \tan^{-1}(\dfrac{590}{333})\\\Rightarrow \theta = 60.56^\circ$

Hence, the rope should be at angle $60.56^\circ$ south of west.