Question

A 2.0 kg block sits on top of an 8.0 kg block. A force of magnitude 17 N is applied to the top block at an angle of 30 degrees from the horizontal . The coefficient of static friction between the two blocks is 0.50 and the lower block is glued to the floor. Does the upper block slip with respect to the lower block?

Answer 1

Answer:Yes

Explanation:

Given

mass $m_1 =2 kg$

mass $m_2=8 kg$

Force of magnitude =17 N

coefficient of static friction $\mu =0.50$

Lower block is glued to floor

Force cos component will try to move the upper block while sin component try to lower the Normal reaction

Normal reaction

$N=mg-F\sin 30 =2\times 9.8-8.5=11.1 N$

and friction force $f_r=\mu N=0.5\times 11.1=5.55 N$

And  force cos component =$F\cos 30 =14.72 N$

As force cos component is more than the friction force therefore it slips w.r.t to lower block