Question

Two identical boxes are being pulled across a horizontal floor at a constant velocity by a horizontal pulling force of 176 N that is applied to one of the boxes, as the drawing shows. There is kinetic friction between each box and the floor. Find the tension in the rope between the boxes.

Answer 1

Answer:

88 N

Explanation:

Since the two boxes are identical, they have the same normal force and kinetic friction constant.

Also, since you know they are moving at a constant velocity, the force of 176 N being applied to one of the boxes must equal the sum of the force of friction from both boxes, that is:

176 = 2 * (F_n * u_k) where F_n is normal force and u_k is coefficient of kinetic friction.

Since the rope between the boxes only has to exactly cancel out 1 box worth of kinetic friction, you get:

F_t = (F_n * u_k) where F_t equals the tension in the rope.

Substitution F_t in the first equation leads to:

176 = 2 * (F_t)

F_t = 88

So the tension in the rope is exactly half of the 176 N force being applied, so it is 88 N.

Answer 2

Answer:

the tension in the rope between the boxes is equal to 88 N

Explanation:

given,

the force applied on one body F = 176 N

When two bodies are moving on horizontal plane at constant velocity then their kinetic friction (f k) is equal to applied force F

According to newton third law the resultant force acting on one body is equal to the resultant force acting on the another body.

T is the tension in the rope

$T- f_k = - (T- f_k)$

T - F = - (T - F)

T - 176 = - (T - 0)

2 T = 176

T = 176/2 = 88 N

so, the tension in the rope between the boxes is equal to 88 N