Video: Finding the Limiting Friction Force Applied to a Brick on a Rough Horizontal Surface

A brick is initially at rest on top of a rough horizontal surface, as shown in the diagram. The brick is pushed by a force 𝐹 parallel to the surface. The weight of the brick |𝑊| = 30 N. The coefficient of static friction of the brick with the surface is 0.75. What must the magnitude of 𝐹 exceed to start the brick moving across the surface?

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Video Transcript

A brick is initially at rest on top of a rough horizontal surface, as shown in the diagram. The brick is pushed by a force 𝐹 parallel to the surface. The weight of the brick equals 30 newtons. The coefficient of static friction of the brick with the surface is 0.75. What must the magnitude of 𝐹 exceed to start the brick moving across the surface?

Okay, so we have this brick initially at rest on a rough horizontal surface. And we see that this brick is being pushed by a force 𝐹 parallel to the surface. For small enough values of 𝐹, the brick actually won’t move. And that’s due to frictional effects between the brick and the surface, leading to a frictional force on the brick. This frictional force is in opposition to the applied force 𝐹. But the frictional force has some maximum value it can achieve. If the applied force 𝐹 exceeds that maximum value, then the brick will start to move.

To figure out the greatest possible static frictional force that acts on the brick, we can recall that the magnitude of the frictional force acting on some object on a flat surface is equal to the object’s weight multiplied by the coefficient of static friction between the object and the surface it rests on. We can say then that the maximum static frictional force acting on the brick is equal to its weight, here we have the magnitude of its weight, multiplied by the coefficient of static friction between the brick and the rough surface.

Just as a side note, we say that this is the maximum static frictional force because the force actually could be less than this. When the force 𝐹 that we apply to the brick is very small, yet increasing, the static frictional force on the brick keeping it in place increases with 𝐹. But as we’ve said, this happens only up to the point where that applied force is either less than or equal to the weight of this brick multiplied by the coefficient of static friction.

When 𝐹 exceeds this value, it’s at that exact moment that the brick will start to move. And that’s why we’re solving for 𝐹 sub f max. Substituting in our given values, this is equal to 30 newtons times 0.75. And if we keep just two significant figures from our answer, that ends up as 23 newtons. Once our applied force 𝐹 exceeds this value, the static frictional force will no longer be able to keep up, so to speak, and the brick will start to move.

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