Question

A 60kg block is at rest on a ramp inclined at 35 degrees.

Answer 1

Answer:

Explanation:

(a)Gravitational Force is W=mg

Frictional force is f

Normal force is N

(b)Since the block is in Equilibrium therefore from diagram we can see that  

$f=mg\sin \theta$

$f=60\times 9.8\times \sin 35$

$f=337.26 N$

(c)Magnitude of Normal Force

$N=mg\cos \theta =481.66 N$

(d)friction is also given by

$f=\mu mg\cos \theta$

and $f=mg\sin \theta$

thus $tan \theta =\mu$

$\mu =0.7$