Question

Two boxes connected by a light horizontal rope are on a horizontal surface, as shown in the following figure. The coefficient of kinetic friction between each box and the surface is 0.30. One box (box B) has mass 4.84 kg , and the other box (box A) has mass m . A force F with magnitude 37.3 N and direction 53.1 ∘ above the horizontal is applied to the 4.84 kg box, and both boxes move to the right with an acceleration of a = 1.50 m/s2 .(Figure 1)

What is the tension T in the rope that connects the boxes?

What is the mass m of the second box? Use the unrounded intermediate answers to calculate the final one.

Answer 1

Balancing forces in vertical on B,

N + 37.3 sin 53.1 degree - 4.84*9.8 = 0

N = -37.3 sin 53.1 degree +4.84*9.8 = 17.6 N

Friction = uN = 17.6*0.3 = 5.28 N

Now in horizontal,

37.3 cos 53.1 degree - 5.28 -T = 4.84*1.50

Tension T = 37.3 cos 53.1 degree - 5.28 - 4.84*1.50

   = 9.856 N

Applying Newtons law on second mass,

T - umg = ma

9.856 = m(a+ug)

m = 9.856/[a+ug)

    = 9.856/[1.5+0.3*9.8]

= 2.22 kg