Question

In Fig.(a) The coefficient of friction is 0.20 between the rope and the fixed drum, and between other surface of contact µ = 0.3. Determine the minimum weight W to prevent downward motion of the 1000 N body.

Answer 1

Since 1000 N weight is on the verge of sliding downwards the rope connecting it is the tight side and the rope connecting W is the slack side. Free body diagrams for W and 1000 N body are shown in Fig. (b).

cos α = 0.8 

sin α = 0.6 

Consider the equilibrium of weight W, 

Σ Forces perpendicular to the plane = 0, gives 

N₁ = W cos α 

N₁ = 0.8 W ...(1) 

∴ F₁ = µN₁ = 0.3 × 0.8 W 

F₁ = 0.24 W ...(2) 

Σ Forces parallel to the plane = 0, gives

T₁ = F₁ + W sin α = 0.24 W + 0.6 W 

= 0.84 W ...(3)

Angle of contact of rope with the pulley = 180° = π radians 

Applying friction equation, we get

Substituting the value of T₁ from (3) 

T₂ = 2.156 W ...(4) 

Now, consider 1000 N body, 

Σ forces perpendicular to the plane = 0, gives 

N₂ – N₁ – 1000 cos α = 0 

Substituting the value of N₁ from (1),

N₂ = 0.8 W + 1000 × 0.8 = 0.8 W + 800 

∴ F₂ = 0.3 N₂ = 0.24 W + 240 ...(5) 

Σ forces parallel to the plane = 0, gives 

F₁ + F₂ – 1000 sin α + T₂ = 0

Substituting the values from (2), (4) and (5), 

0.24 W + 0.24 W + 240 – 1000 × 0.6 + 2.156 

W = 0 W = 136.57 N.