Question

58. The system shown is just on the verge of slipping. The co-efficient of static friction between the block and the
table top is
WIT
W' = 40N
30°
W= 8N
(1) 0.5
(2) 0.95
(3) 0.15
(4) 0.35​

Answer 1

Equilibrium at block R

T1= 8N

T2sin30=T1

T2 = 16N

T2cos30=T3

Equilibrium at Block Q

N=40N

T3= f

T3=μN

μ=T2cos30/40

μ=0.35

Answer 2

Answer:

Explanation:

Tension in the string attached to the wall be T, coefficient of friction between block of weight 40N and the floor below it be 'u'.

Then resolving tension in vertical and horizontal directions, we get Tcos(30) = u*(normal force acting on the block with weight 40N)

and Tsin(30) = weight of the other block = 8N

so, T = 16 and u = 16*cos(30) / (40) = 0.35