Question

32.
A block of mass 4 kg is placed over a rough inclined plane as shown in the figure. The coefficient of
friction between the block and the plane is u=0.6. A force F is applied on the block at an angle of 30°.
The contact force between the block and the plane is
450
2016 M​

Answer 1

Answer:

Drawing free - body diagram of block , we get

N+Fsin30∘=mgcos37∘

N=mgcos37∘−Fsin30∘

=(4)(10)(45)−(10)(12)=27N…(i)

fmax=μN=0.5×27=13.5N

mgsin37∘=(4)(10)(35)=24N

andFcos30∘=(10)(3–√5)=53–√N

Now since mgsin37∘>fmax=Fcos30∘

Therefore , block will slide down and friction will be maximum .Therefore net contact force is

Fc=N2+(fmax)2−−−−−−−−−−−√

=(27)2+(13.5)2−−−−−−−−−−−−√=13.55–√N