Question

Part 1 :
A gust of wind blows an apple from a tree. As the apple falls, the force of gravity on the apple is 9.42N downward, and the force of the wind on the apple is 1.68N to the right.
What is the magnitude of the net external force on the apple? Answer in units of N.
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Part 2 :
What is the direction of the net external force on the apple (measured from the downward vertical, so that the angle to the right of downward is positive)? Answer in units of degrees.

Answer 1

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

$F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N$

2) $10^{\circ}$

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

$\theta = tan^{-1}(\frac{F_x}{F_y})$

where

$F_x$ is the force in the horizontal direction

$F_y$ is the force in the vertical direction

In this problem,

$F_x = F_w = 1.68 N$

$F_y = W = 9.42 N$

and so we find:

$\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}$