Solution | Q.29, Civil GATE 2021, Set1 | Wedge M and Block N Subjected to forces P and Q.

Problem Statement:

A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction (μ) along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is 30°.

The limiting force P, in terms of Q, required for the impending motion of block N to just move it in the upward direction is given as P = αQ . The value of the

coefficient ‘α’ (round off to one decimal place) is

Solution: (D) 0.9

Reaction Forces and the frictional forces are named appropriately, as shown in the FBDs.

Friction Force, F_f = μ*R_n1 = 0.20R_n1 —(1)

Also Check – GATE 2021 | Q.28 Solution

For the overall system’s FBD, ∑F_y =0, hence R_n = Q —(2)

Consider the FBD of the wedge M,

As ∑F_x = 0 => P = R_n1*Sin(30˚) + F_f*Cos(30˚)

or P = 0.5R_n1 + 0.20*R_n1*Cos(30˚) or R_n1 = 1.485P —(3)

Also, ∑F_y = 0, => R_n = R_n1*Cos(30˚) – F_f*Sin(30˚)

Put values from eq(1), eq(2) and eq(3)

or Q = 1.485P*Cos(30˚) – 0.20*1.485P*Sin(30˚)

or Q = 1.137P

or P = 0.878Q ≈ 0.9P

hence α ≈ 0.9

Also Check: Solution of GATE 2021, CE, Set1 (Q.1 to Q.100)