Two masses are connected by a string which goes over an ideal pulley

Asked by maham237 @ 23/12/2020 in Physics viewed by 315 People

Two masses are connected by a string which goes over an ideal pulley as shown in Fig. 5-6. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0° to the horizontal. The coefficient of kinetic friction between block A and the plane is 0.400. Block B has a mass of 2.77 kg. What is the acceleration of the blocks?

Answered by maham237 @ 23/12/2020

Answer:

The acceleration of the blocks is 0.392 m/s²

Explanation:

Lets revise the newton second low

Newton's second law of motion states that an object will accelerate

when there is a net external force acting on the object

∑ force in direction of motion = mass × acceleration

The system will move by same acceleration

So we will take all the forces acting on blocks to find the acceleration

Assume that the system will move up the plane because the

weight of block B is greater than the component of wight of

block A

At first we must distribute the mass of block A to two components :

One parallel to the plane ⇒ (m1)g sin 30°

One perpendicular to the plane ⇒ (m1)g cos 30°

The forces acting in block A are:

1. Tension (T) acting up in the rope

2. Frictional force (F) acting down (opposite to the direction of motion)

3. The parallel component of the weight acting down (mg sin 30)

According to Newton Low:

T - (m1)g sin(30) - F = (m1)a ⇒ (1), where (m1) is the mass of block A and

"a" is the acceleration of the system

Block b will move up

The force acting on block B are:

1. Tension (T) upward

2. Weight of block B (m2)g downward

According to Newton Low:

(m2)g - T = (m2)a ⇒ (2), where m2 is the mass of block B

Add equations (1) and (2) to eliminate the tension

(m2)g - (m1)g sin(30) - F = (m1 + m2)a ⇒ (3)

Frictional force F = μR, where R is the normal reaction on the block A

R = (m1)g cos30

F = μ (m1) cos30

μ = 0.4 , m1 = 3 , g = 9.8 m/s²

F = (0.4)(3)(9.8) cos30 = 10.184

m2 = 2.77

Substitute the vales in equation (3)

(2.77)(9.8) - (3)(9.8) sin30 - 10.184 = (3 + 2.77)a

27.146 - 14.7 - 10.184 = 5.77 a

2.262 = 5.77 a

Divide both sides by 5.77

a = 0.392 m/s²

The acceleration of the blocks is 0.392 m/s²

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