Question Video: Finding the Magnitude of the Force Pulling a Body on a Rough Horizontal Plane given the Time and Velocity Mathematics

A body of 12 kg was placed on a rough horizontal plane. It was pulled by a force whose line of action made an angle of πœƒ upward of the plane, where sin πœƒ = 3/5. If the body starting from rest moved a distance of 804 cm in 4 seconds and the coefficient of friction was 3/4, find the magnitude of the pulling force. Take the acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

A body of 12 kilograms was placed on a rough horizontal plane. It was pulled by a force whose line of action made an angle of πœƒ upward of the plane, where sin πœƒ is equal to three-fifths. If the body starting from rest moved a distance of 804 centimeters in a four seconds and the coefficient of friction was three-quarters, find the magnitude of the pulling force. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per second squared.

We will begin by sketching a diagram modeling the scenario in this question. We are told that a body of mass 12 kilograms was placed on a rough horizontal plane. This will exert a downward force of 12𝑔, where 𝑔 is equal to 9.8 meters per second squared. Using Newton’s third law, there will be a normal reaction force acting in the opposite direction, in this case acting vertically upwards. The body is pulled by a force whose line of action makes an angle of πœƒ upward of the plane as shown. We are also told that sin πœƒ is equal to three-fifths. We will be able to find the horizontal and vertical components of this force using our knowledge of right-angled trigonometry.

As the plane is rough, there will be a frictional force 𝐹 π‘Ÿ acting against the motion. And since the coefficient of friction is three-quarters, we can find an expression for this using the formula 𝐹 π‘Ÿ is equal to πœ‡ multiplied by 𝑅. The frictional force will be equal to three-quarters the normal reaction force. Our aim in this question is to find the magnitude of the pulling force 𝐹. So we will begin by resolving vertically and horizontally.

As already mentioned, we can work out the horizontal and vertical components of this force by using our knowledge of right angle trigonometry. In the right triangle drawn, the horizontal component is adjacent to angle πœƒ and the vertical component is opposite angle πœƒ. The force 𝐹 is the hypotenuse of our triangle. We know that the sin of angle πœƒ is equal to the opposite over the hypotenuse and the cos of angle πœƒ is the adjacent over the hypotenuse. As sin πœƒ is equal to three-fifths, we have three-fifths is equal to 𝑦 over 𝐹. And multiplying through by 𝐹, we have 𝑦 is equal to three-fifths 𝐹. The vertical component of our force is three-fifths 𝐹 and acts vertically upwards.

Using our knowledge of the three-four-five Pythagorean triples, we know that if sin πœƒ is three-fifths, cos πœƒ is four-fifths, and this must be equal to π‘₯ over 𝐹. Once again, we can multiply through by 𝐹 such that the horizontal component of the force is four-fifths 𝐹. We are now in a position to resolve horizontally and vertically using Newton’s second law 𝐹 equals π‘šπ‘Ž. The sum of our forces is equal to the mass multiplied by the acceleration.

In the horizontal direction, we have two forces, four-fifths 𝐹 and the frictional force 𝐹 π‘Ÿ. Taking the positive direction to be the direction of motion, the sum of our forces is four-fifths 𝐹 minus 𝐹 π‘Ÿ. This is equal to the mass of 12 kilogram multiplied by the acceleration π‘Ž, which is as yet unknown. We can replace the frictional force 𝐹 π‘Ÿ with three-quarters multiplied by the normal reaction force.

Resolving vertically where the positive direction is vertically upwards, we have 𝑅 plus three-fifths 𝐹 minus 12𝑔. As the body is not accelerating in this direction, this is equal to zero. As 12 multiplied by 9.8 is 117.6, this simplifies to 𝑅 plus three-fifths 𝐹 minus 117.6 is equal to zero. We now have two equations but with three unknowns. We can calculate the value of the acceleration π‘Ž using some extra information from the question that we have not used so far.

We are told that the body starts from rest and moves a distance of 804 centimeters in four seconds. We can therefore use the equations of motion or SUVAT equations to calculate the acceleration π‘Ž. The body starts from rest, so 𝑒 is equal to zero meters per second. It travels a distance of 804 centimeters or 8.04 meters in four seconds. Therefore, 𝑠 is equal to 8.04 and 𝑑 equals four. We will use the equation 𝑠 is equal to 𝑒𝑑 plus a half π‘Žπ‘‘ squared.

Substituting in our values, we have 8.04 is equal to zero multiplied by four plus a half multiplied by π‘Ž multiplied by four squared. The right-hand side simplifies to eight π‘Ž. We can then divide through by eight, giving us π‘Ž is equal to 201 over 200, which is equal to 1.005 meters per second squared. We can substitute this value into equation one so that this equation simplifies to four-fifths 𝐹 minus three-quarters 𝑅 is equal to 12.06.

We now have a pair of simultaneous equations that we can solve to calculate the value of 𝐹. One way of doing this is by substitution. Subtracting three-fifths 𝐹 and adding 117.6 to both sides of equation two gives us 𝑅 is equal to 117.6 minus three-fifths 𝐹. We can then substitute this expression for 𝑅 into equation one. This gives us four-fifths 𝐹 minus three-quarters multiplied by 117.6 minus three-fifths 𝐹 is equal to 12.06.

Distributing the parentheses, we have four-fifths 𝐹 minus 88.2 plus nine twentieths 𝐹 is equal to 12.06. We can then collect like terms on the left-hand side and add 88.2 to both sides. Four-fifths 𝐹 plus nine twentieths 𝐹 is equal to five-quarters or 1.25𝐹. This is equal to 100.26. We can then divide through by 1.25 such that 𝐹 is equal to 80.208.

The magnitude of the pulling force 𝐹 is therefore equal to 80.208 newtons.

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