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?

Answer 1

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is $5 N$

Tray makes an angle of $\theta =12^{\circ}$

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force $W_p=W\cos \theta$

$W_p=5\times \cos 12=4.89 N$

Along the Tray

$W_{along}=W\sin \theta$

$W_{along}=5\times \sin 12=1.04 N$

Thus 1.04 N is the magnitude of force that will cause Platter to slide down