Question Video: Determining the Magnitude of Force of the Motion of a Body on a Rough Horizontal Plane with Friction Mathematics

A body of mass 440 g was resting on a rough horizontal plane whose coefficient of friction was 1/4. The body was dragged across the plane by the action of a horizontal force. Given that the action of this force resulted in a uniform acceleration of 170 cm/sΒ², find the magnitude of the force. Take the acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

A body of mass 440 grams was resting on a rough horizontal plane whose coefficient of friction was one-quarter. The body was dragged across the plane by the action of a horizontal force. Given that the action of this force resulted in a uniform acceleration of 170 centimeters per second squared, find the magnitude of the force. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per second squared.

We will begin by sketching a diagram to model the scenario in this question. The body has a mass of 440 grams. This will therefore exert a downward force of mass multiplied by gravity. If the mass is in kilograms and the acceleration due to gravity in meters per second squared, the force will be in newtons. Alternatively, if our mass is measured in grams and acceleration in centimeters per second squared, the force will be in dynes. This means that we need to either convert the 440 grams to kilograms or the 9.8 meters per second squared to centimeters per second squared.

In this question, we will let 𝑔 be 980 centimeters per second squared as there are 100 centimeters in a meter. The downward force is therefore equal to 440 multiplied by 980 dynes. Typing this into our calculator gives us 431,200. By recalling Newton’s third law, we know there will be a normal reaction force 𝑅 acting in the opposite direction to this. We are told that there is a horizontal force 𝐹 acting on the body and it is the magnitude of this force that we are trying to calculate.

Since the plane is rough, there will be a frictional force 𝐹 r acting against the motion. We know that the coefficient of friction πœ‡ is equal to one-quarter. As the body is moving, we know that the frictional force will be at its maximum. And when this is the case, it is equal to πœ‡ multiplied by the normal reaction force 𝑅. The frictional force is therefore equal to one-quarter of the normal reaction force.

The final piece of information we are given in the question is that the body is moving with a uniform acceleration of 170 centimeters per second squared. In order to calculate the magnitude of the force 𝐹, we will use Newton’s second law. This states that 𝐹 equals π‘šπ‘Ž. The sum of the forces is equal to the mass multiplied by the acceleration.

Resolving vertically, we have the normal reaction force and the weight force. If we let the positive direction be vertically upwards, the sum of our forces is 𝑅 minus 431,200. As the body is not accelerating in this direction, this is equal to zero. Adding 431,200 to both sides of this equation, we have 𝑅 is equal to 431,200 dynes. Resolving horizontally where the positive direction is the direction of travel, we have 𝐹 minus 𝐹 r. This is equal to the mass of 440 grams multiplied by the acceleration of 170 centimeters per second squared.

Recalling that the frictional force 𝐹 r is equal to a quarter of the normal reaction force 𝑅, then the left-hand side simplifies to 𝐹 minus 107,800. Multiplying 440 by 170 gives us 74,800. We can then add 107,800 to both sides, giving us 𝐹 is equal to 182,600. The magnitude of the force 𝐹 is therefore equal to 182,600 dynes.

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