A 50 kg box is at rest on a horizontal surface. When an 80 N force is applied to the box, it slides on the surface with a constant velocity.

How will the force of friction change if an identical box is placed on top of the 50 kg box?

As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box So we find Mu. Mu = coefficient of friction between box and horizontal surface = Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is ** = Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons**

Answered by Everest2017 | 2024-06-10

When an identical box is stacked on the 50 kg box, the force of friction increases due to the increase in the normal force, leading to a greater maximum potential friction force. The total weight on the lower box becomes 100 kg, raising the normal force to 981 N. Since the normal force is directly related to friction, we conclude that the force of friction will increase in this scenario.
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Answered by Everest2017 | 2024-12-18