Question Video: Static and Kinetic Friction on a Rough Inclined Plane | Nagwa Question Video: Static and Kinetic Friction on a Rough Inclined Plane | Nagwa

# Question Video: Static and Kinetic Friction on a Rough Inclined Plane Mathematics • Third Year of Secondary School

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A body of weight 24 N is placed on a rough plane inclined at 30° to the horizontal. The coefficient of static friction between the body and the plane is 1/6√3, and the coefficient of kinetic friction is 1/12√3. A force is pushing the body upward in the direction of the line of greatest slope. Determine the magnitude of the force that would make the body about to move.

04:24

### Video Transcript

A body of weight 24 newtons is placed on a rough plane inclined at 30 degrees to the horizontal. The coefficient of static friction between the body and the plane is one over six root three. And the coefficient of kinetic friction is one over 12 root three. A force is pushing the body upward in the direction of the line of greatest slope. Determine the magnitude of the force that would make the body about to move.

Sketching in this scenario, it could look like this where our body is at rest on an inclined plane of 30 degrees. Situated like this, our body is subject to a static frictional force, which depends on the coefficient of static friction given as one over six root three. We’re told that there’s a force, we’ll call it 𝐹, that acts on the body in the direction of the line of greatest slope. And we want to figure out how big that force could be, such that the body would be about to move.

Let’s first give ourselves some space on screen to work. And then let’s include in our sketch all of the forces that are acting on our body. We’re told that there’s a weight force of magnitude 24 newtons. And we also know that there’s a normal force perpendicular to the surface of our slope. Along with all this, there is a static frictional force. Now, if we had no applied force, what we’ve called 𝐹, pushing our body up the incline, then our static frictional force would point in that same direction. This would be so it would resist the block sliding down the plane. But in our scenario, we want to solve for the maximum possible value of 𝐹 such that the block is just barely not moving up the incline. That means that, here, the static frictional force is actually acting down the incline.

As we begin analyzing these forces acting on our body, we can orient ourselves with a pair of orthogonal axes. We’ll say that forces in motion acting up the incline are positive, and those acting the opposite way are negative. Likewise, forces in motion perpendicularly away from our surface are positive, and those in the other direction are negative. Knowing this, we can apply Newton’s second law of motion to both of what we’ll call our 𝑦-direction and our 𝑥-direction.

Looking first in the 𝑥-direction, we see that our force 𝐹 acts in the positive 𝑥, while our frictional force acts opposite this. These, however, aren’t the only 𝑥-direction forces. That’s because our weight force has a component right here in the negative 𝑥-direction. Now, this angle right here in this triangle is equal to 30 degrees. And since this intersection is a right angle, we can see that this 𝑥-component of our weight force is 24 newtons times the sin of 30 degrees. The sin of 30 degrees we can recall is one-half, so we’ll substitute that in right away.

So these then are indeed all the forces acting on what we’ve called our 𝑥-direction. By Newton’s second law, these add up to our object’s mass times acceleration in this dimension. But remember that we’re solving for 𝐹 such that our object is not yet in motion. In other words, 𝑎 sub 𝑥 is zero. That means our whole right side is zero, which means we can rearrange and see that the maximum force we can apply to our body without it yet being put in motion is equal to the static frictional force plus 12 newtons. In general, the force of static friction is equal to the coefficient of static friction times the normal force acting on the relevant body.

In our scenario, we know 𝜇 sub 𝑠. But what about 𝐹 sub 𝑁? To figure that out, we’ll want to look at Newton’s second law of motion applied in the 𝑦-direction. Here, we have a positive, normal force but then a negative component of our weight force here on our sketch. This component is 24 newtons times the cos of 30 degrees where the cos of that angle equals the square root of three over two. Just like in the 𝑥-direction, because there’s no acceleration in the 𝑦-direction, these forces add up to zero. This tells us that the normal force equals 12 times the square root of three newtons.

Now that we know 𝐹 sub 𝑁, we can substitute it as well as 𝜇 sub 𝑠 into our equation for the frictional force. And multiplying these quantities together, we see that a square root of three cancels out and that the maximum frictional force, therefore, is two newtons. 𝐹 then is 14 newtons. This is the magnitude of the largest force that could act directly up the incline without our body being put in motion.

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