Force — Question 2

Question 2

Perpendicular forces add as vectors

Two forces on one point are 6 N east and 8 N north.

1. Find resultant magnitude and direction north of east.

2. Find the single force that would balance them.

3. Explain why their resultant magnitude is not 14 N.

Question 2 - Solution

Perpendicular forces add as vectors

(a) Use components.

$$\vec R=6\hat i+8\hat j,\quad\boxed{|\vec R|=\sqrt{6^2+8^2}=10\,\mathrm N},\quad\theta=53.130^\circ$$ $\vec F=(6,8)$ N, so $|\vec F|=10$ N and $\theta=\tan^{-1}(8/6)=53.130^\circ$ north of east.

(b) Negate the full vector.

$$\vec F_{\mathrm{balance}}=-\vec R=(-6,-8)\,\mathrm N,\quad\theta_b=233.130^\circ$$ The balancing force is $(-6,-8)$ N, magnitude 10 N and direction $233.130^\circ$ measured counterclockwise from east.

(c) Retain relative directions.

$$R^2=6^2+8^2+2(6)(8)\cos90^\circ=100\,\mathrm{N^2}$$ Adding magnitudes gives 14 N only for parallel forces in the same direction. Here the perpendicular components form a right triangle, so Pythagoras determines the magnitude.