Question

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitational force on the platter is 5.0 N, what is the magnitude of the force parallel to the tray that tends to cause the platter to slide down the tray? (Disregard friction.)

Answer 1

Answer:1.039 N

Explanation:

Given

inclination of tray$=12^{\circ}$

gravitational Force=5 N

Now this gravitational force has two component i.e.

$5\sin \theta$ is parallel to the tray =1.039 N

$5\cos \theta$ is perpendicular to the tray =4.890 N