A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6 What should be the minimum value of force P, such that the block doesnot move downward? (take `g=10ms^(-2)`)

Options

A. 23N
B. 32N
C. 25N
D. 18N

Answer

B

Explanation

Since `f_(i)=mu N=0.6x100xx1/(sqrt(2))=60/(sqrt(2))`

and Net downward force is `3+100/(sqrt(2))`
So to avoid downward slipping to block, we need to apply P upwards (i.e. up along the incline).
`:.=(3sqrt(2)+40)/(sqrt(2))=(6+40sqrt(2))/2=3+20sqrt(2)=3+20(1.414)`
`=31.28~~32N`.