Refer to figure.If the co-efficient of friction is 0.60 for all contact surfaces and θ = 30o,what force P applied to the block B acting down and parallel to the incline will start motion and what will be the tension in the cord parallel to inclined plane attached to A.

Take WA=120 N and WB=200 N.

#### Solution

Given : : μ=0.6

θ = 30°

W_{A} = 120 N

W_{B} = 200 N

To find : Force P

Solution :

F_{1} = μN_{1} = 0.6N_{1} ………(1)

F_{2} = μN_{2} = 0.6N_{2} ……….(2)

Consider FBD of block A

The block is considered to be in equilibrium

Applying conditions of equilibrium

ΣFy = 0

N_{1} - 120cos30 = 0

N_{1} = 103.923 N ……….(3)

From (1)

F_{1} = 0.6 x 103.923

= 62.3538 N

Applying conditions of equilibrium

ΣFx = 0

F_{1} + 120sin30 – T = 0

T = 122.3538 N**Consider FBD of block B ****Applying conditions of equilibrium**ΣFy = 0

N

_{2}- N

_{1}- 200cos30 = 0

F

_{2}= 0.6 x 277.1281

= 166.2769 N From (2)

Applying conditions of equilibrium

ΣFx = 0

P - F

_{1}- F

_{2}+ 200sin30 = 0

P = 128.6307 N

Force required on block B to start the motion is 128.6307 N Tension T in the cord parallel to inclined plane attached to A=122.3538 N