A block of mass m = 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 μ=0.5. A force F = 10 N is applied on the block at an angle of 30∘. Find the contact force (in N) between the block and the plane.

Drawing free body diagram of block.⇒ N+Fsin⁡30∘=mgcos⁡37∘or, N=mgcos⁡37∘−Fsin⁡30∘=(4)(10)45−(10)12=27N             …(i)fmax=μN=0.5×27=13.5Nmgsin⁡37∘=(4)(10)35=24Nand Fcos⁡30∘=(10)32=53NNow since mgsin⁡37∘>fmax+Fcos⁡30∘Therefore, block will slide down and friction will be maximum.Therefore, net contact force isFC=N2+fmax2=(27)2+(13.5)2=13.55N=30.2N