Question

In the diagram shown, the blocks A, B and C weight 3 kg, 4 kg and 5 kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall while B and C are connected by a light flexible cord passing round a frictionless pulley. The force F necessary to drag C along the horizontal surface to the left at constant speed is (Assume that the arrangement shown in the diagram, B on C and A on B is maintained all through and g=10 m/s2)

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Solution

Given,

mass of block A, MA=3 kg,

mass of block B, MB=4 kg

mass of block C, MC=5 kg

μ=0.25 [ any two surfaces ]

When force F is applied on C, the block C will move towards left.

The free body diagram for C:

As C is moving with constant speed

F=f1+f2+T.....(1)

Where f1 is friction between floor and block C and f2 between blocks B and C.

FBD for mass B:

As B moves with constant speed

T=f2+f3.....(2)

From equation (1) and (2), we get

F=f1+f2+f2+f3

⇒F=f1+2f2+f3

Substituting the values of f1, f2 and f3, we get

⇒F=f1+2f2+f3

⇒F=μ(MA+MB+MC)g+μ(MA+MB)g+μ(MAg)

Substituting the values, we get

⇒F=0.25×(3+4+5)10+0.25×2(3+4)10+0.25×(3)×10

∴F=72.5 N

Accepted answers : 72.5 , 72.50

mass of block A, MA=3 kg,

mass of block B, MB=4 kg

mass of block C, MC=5 kg

μ=0.25 [ any two surfaces ]

When force F is applied on C, the block C will move towards left.

The free body diagram for C:

As C is moving with constant speed

F=f1+f2+T.....(1)

Where f1 is friction between floor and block C and f2 between blocks B and C.

FBD for mass B:

As B moves with constant speed

T=f2+f3.....(2)

From equation (1) and (2), we get

F=f1+f2+f2+f3

⇒F=f1+2f2+f3

Substituting the values of f1, f2 and f3, we get

⇒F=f1+2f2+f3

⇒F=μ(MA+MB+MC)g+μ(MA+MB)g+μ(MAg)

Substituting the values, we get

⇒F=0.25×(3+4+5)10+0.25×2(3+4)10+0.25×(3)×10

∴F=72.5 N

Accepted answers : 72.5 , 72.50

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