Answer to Question #132409 in Mechanics | Relativity for K.Arun sriram

Question

determine the horizontal force P required to start moving the 500N block up the inclined surface.assume static friction is 0.3 and angle is 30

Answer 1

As per the question

weight of block ,W=500N

Angle on inclination, $\theta=30\degree$

cofficient of friction, $u_s=0.3$

As the block is moving up the inclined surface ,therefore the equation became

$pcos\theta=Wsin\theta+f_s$

$pcos\theta=Wsin\theta+u_sR$

Where R is the Normal reaction to the block from inclined plane

$pcos30=500sin30+0.3(Wcos\theta+psin\theta)$

$0.866p=250+0.3(433+psin30)$

0.866p=250+$0.3\times433+0.3\times p\times Sin30$

$0.866p=250+129.9+0.15p$

$0.866p-0.15p=379.9$

0.716p=379.9

P=$\frac{379.9}{0.716}$

=530.58N