Question

Block A of weight 2000N is kept on the inclined plane at 35° .It is connected to weight B by an inextensible string passing over smooth pulley.
Determine the weight of pan B so that B just moves down.Assume μ=0.2.

Given : Weight of block A=2000N
Angle of inclined plane = 35°
μ=0.2
To find : Weight of pan B 

 

Answer 1

The pan B is in equilibrium
Applying the conditions of equilibrium

ΣF_Y=0
T-WB=0
T=WB ………..(1)
Applying the conditions of equilibrium on block A

ΣF_Y=0
N-WAcos35+Tsin20=0
From (1)
N=2000cos35-WBsin20 ………..(2)
F_S=μ_sN
F_S=0.2(2000cos35-WBsin20)
F_S=400cos35-0.2WBsin20
Applying the conditions of equilibrium on block A

ΣF_X=0
Tcos20-W_Asin35-F_S=0
W_Bcos20-2000sin35-(400cos35-0.2WBsin20)=0(From 1 and 2)
`W_B=(2000sin35+400cos35)/(cos20+0.2sin20)`
W_B=1462.9685 N

The weight of pan B so that pan B just moves down is 1462.9685 N