Two identical boxes of masses 45kg both experience a sliding force of U=0.15. Find the tension in the chord and the acceleration of the boxes

Answer 1

To find the tension in the chord, we need to consider the forces acting on the boxes.

Let T be the tension in the chord and a be the acceleration of the boxes.

The net force acting on each box is given by the force of friction, which is the product of the coefficient of friction and the normal force:
F_friction = U * (mass * gravity)
F_friction = 0.15 * (45kg * 9.8m/s^2)
F_friction = 66.15 N

The tension in the chord will be equal to the force of friction since there are no other horizontal forces acting on the boxes:
T = F_friction
T = 66.15 N

Now, we can find the acceleration of the boxes using Newton's second law:
ΣF = mass * acceleration
T - F_friction = mass * acceleration
66.15 - 66.15 = 45kg * a
0 = 45kg * a
a = 0 m/s^2

Therefore, the tension in the chord is 66.15 N and the acceleration of the boxes is 0 m/s^2.