# Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62 degree north of west. What is the magnitude of the body's acceleration?

Jun 30, 2018

$a \cong 3.8 \frac{m}{s} ^ 2$

#### Explanation:

Let us work with an orthogonal reference system with East being positive along one axis and North being positive along the other axis.

The first force is $9.0 N$ east
The other force, converted to components along our 2 axes, is $8.0 N \cdot \cos {62}^{\circ} = 3.76 N$ west and $8.0 N \cdot \sin {62}^{\circ} = 7.06 N$ north.

In the east-west direction, the sum of components along that axis is 9.0 N - 3.76 N = 5.24 N east.
In the north-south direction, the sum of components along that axis is 7.06 N north.

Pythagoras will lead us to the magnitude of the resultant force.
#F_"res" = sqrt((9.0 N)^2 + (7.06 N)^2) = 11.4 N

Newton's 2nd Law will lead us to the acceleration of the 3.0 kg body.
$a = {F}_{\text{res}} / m = \frac{11.4 N}{3 k g} = 3.81 \cong 3.8 \frac{m}{s} ^ 2$

I hope this helps,
Steve