Question Video: Finding the Friction Acting on a Block Pulled by an Inclined Force on a Rough Ground Mathematics

Video Transcript

A horse was pulling a wooden block along a section of horizontal ground. The pulling force of the horse was 22 kilogram-weight which was acting on the block at an angle of 60 degrees to the vertical. Given that the block was moving uniformly, determine the magnitude of friction acting on the block.

We will begin by sketching the block and the forces acting on it. We are told that there is a pulling force of 22 kilogram-weight acting at an angle of 60 degrees to the vertical, as shown. We are told that the block was moving uniformly. Therefore, its acceleration is zero meters per second squared. And we are asked to calculate the magnitude of the friction acting on the block; we will call this 𝐅 𝑟. And we know that this will act horizontally in the direction against the motion.

In order to calculate this force, we’ll need to resolve horizontally and as such we’ll need to work out the horizontal component of the 22 kilogram-weight force. One way of doing this is using our knowledge of right angle trigonometry. By drawing a right triangle, we see that the 22 kilogram-weight force is the hypotenuse and the horizontal component of this force is the opposite. We know that the sin of any angle 𝜃 is equal to the opposite over the hypotenuse. This means that the sin of 60 degrees is equal to 𝑥 over 22. We know that the sin of 60 degrees is equal to root three over two. And we can then multiply through by 22, giving us 𝑥 is equal to 11 root three. The horizontal component of the pulling force is 11 root three kilogram-weight.

We can now resolve horizontally using Newton’s second law. This states that the sum of our forces is equal to the mass multiplied by the acceleration. If we let the positive direction be the direction of motion, the sum of our forces is 11 root three minus 𝐅 𝑟. And since we have already established that there is no acceleration, this is equal to zero. We can then add the frictional force 𝐅 𝑟 to both sides of our equation.

The magnitude of the friction acting on the block is therefore equal to 11 root three kilogram-weight.