Two boxes of masses m=35kg and m2=45kg, are hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod. They system starts moving from rest. Assuming that friction between the rod and the rope is negligible, determine the magnitude of

(a) the acceleration of the boxes
(b) the tension in the rope
(c) the magnitude of each box's displacement after 0.5s

Given two boxes of masses m1 = 35kg and m2 = 45kg hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod, we will analyze the forces acting on each body.

According Newton's second law, Force = mass ×acceleration

The forces acting on body of mass m1 are the tension (T) and the frictional force (Ff) which opposes the tension.

Taking the sum of horizontal forces acting on mass m1, we will have;

T +(-Ff) = m1a

T - Ff = m1a... (1)

For the mass m2, the forces acting on the body are in the vertical direction and this forces are the weight (W) acting downwards and the tension(T) acting upwards. The sum of the forces in the body is given as ;

W + (-T) = m2a

W-T = m2a ...(2)

Since W = mg, equation 2 will become;

m2g - T = m2a...(2)

Solving equation 1 and 2 simultaneously to get the tension and the acceleration, we have;

T - Ff = m1a ... 1

m2g - T = m2a ... 2

Since friction is negligible, Ff = 0

Adding the two equation will give;

m2g-Ff = m2a+m1a

Since Ff =0

m2g = (m2+m1)a

a = m2g/m1+m2

a = 45(9.8)/45+35

a =441/80

a = 5.51m/s²

b) Substituting a = 5.51 into equation 1 to get the tension T in the rope will give;

T = m1a

T = 35×5.51

T = 192.94N

c) since velocity = displacement/time

Displacement = velocity × time

To get the velocity, since acceleration = velocity/time,

Velocity = acceleration ×time

Velocity = 5.51× 0.5

Velocity = 2.76m/s

Displacement of each box will be the same since they are moving with the same acceleration.

Displacement = 2.76m/s × 0.5s

Displacement of each boxes = 1.38m