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?

1 Answer
Jun 30, 2018

Answer:

#a ~= 3.8 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*cos62^@ = 3.76 N# west and #8.0 N*sin62^@ = 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_"res"/m = (11.4 N)/(3 kg) = 3.81 ~= 3.8 m/s^2#

I hope this helps,
Steve