Question

Force of friction on connected bodies

When bodies are in contact, there are mutual contact forces satisfying the

third law of motion. The component of contact forces normal to the surfaces

in contact is called normal reaction. The component parallel to the surfaces

in contact is called friction.

In the figure 8 kg and 6 kg are hanging stationary from a rough pulley and

are about to move. They are stationary due to roughness of the pulley.

(i) Which force is acting between pulley and rope?

(a) Gravitational force (b) Tension force (c) Frictional force (d) Buoyant force

(ii) The normal reaction acting on the system is

(a) 8g (b) 6g (c) 2g (d) 14g

(iii) The tension is more on side having mass of

(a) 8kg (b) 6kg (c) Same on both (d) Nothing can be said

(iv) The force of friction acting on the rope is

(a) 20 (b) 30 N (c) 40 N (d) 50 N

(v) Coefficient of friction of the pulley is

(a) 1/6 (b) 1/7 (c) 1/5 (d) 1/4​

Answer 1

Answer:

1=TENSION

2=14G

3=8KG

4=20N

5=1/4

Answer 2

i) The force acting between pulley and rope is the tension force in the rope. Therefore, the correct answer is (b) Tension force.

(ii) The normal reaction acting on the system is equal to the sum of the weights of the masses hanging from the pulley, which is 8g + 6g = 14g. Therefore, the correct answer is (d) 14g.

(iii) The tension force in the rope is the same on both sides of the pulley, since the rope is assumed to be massless and inextensible. Therefore, the correct answer is (c) Same on both.

(iv) The force of friction acting on the rope can be calculated using the formula:

friction force = coefficient of friction × normal reaction

The normal reaction is equal to the weight of the 6 kg mass, which is 6g. To find the coefficient of friction, we need to know the force required to make the masses move. Since the masses are just about to move, the tension force in the rope is equal to the maximum static friction force that can act on the pulley. Therefore, we have:

Tension force = maximum static friction force

Tension force = coefficient of friction × normal reaction

Substituting the values, we get:

40 = coefficient of friction × 6g

coefficient of friction = 40/(6g) = 40/(6 × 9.8) = 0.68 (approx.)

Therefore, the force of friction acting on the rope is:

friction force = coefficient of friction × normal reaction

friction force = 0.68 × 6g

friction force = 40.8 N (approx.)

Therefore, the correct answer is (c) 40 N.

(v) The coefficient of friction of the pulley can be calculated using the formula:

coefficient of friction = maximum static friction force / normal reaction

From the previous calculations, we know that the maximum static friction force is equal to the tension force in the rope, which is 40 N. The normal reaction is equal to the weight of the 6 kg mass, which is 6g. Substituting the values, we get:

coefficient of friction = 40 / (6g)

coefficient of friction = 40 / (6 × 9.8)

coefficient of friction = 0.68 (approx.)

Therefore, the correct answer is (c) 1/5.

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