### 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 upward of the plane, where sin π 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 π 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 12π, where π is equal to 9.8 meters per second squared. Using Newtonβs 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 upward of the plane as shown. We are also told that sin π 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 πΉ π acting against the motion. And since the coefficient of friction is three-quarters, we can find an expression for this using the formula πΉ π is equal to π multiplied by π
. 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 πΉ. 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 and the vertical component is opposite angle π. The force πΉ is the hypotenuse of our triangle. We know that the sin of angle π is equal to the opposite over the hypotenuse and the cos of angle π is the adjacent over the hypotenuse. As sin π is equal to three-fifths, we have three-fifths is equal to π¦ over πΉ. And multiplying through by πΉ, we have π¦ is equal to three-fifths πΉ. The vertical component of our force is three-fifths πΉ and acts vertically upwards.

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

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

Resolving vertically where the positive direction is vertically upwards, we have π
plus three-fifths πΉ minus 12π. 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 π
plus three-fifths πΉ 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 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. The body starts from rest, so π’ is equal to zero meters per second. It travels a distance of 804 centimeters or 8.04 meters in four seconds. Therefore, π is equal to 8.04 and π‘ equals four. We will use the equation π is equal to π’π‘ plus a half ππ‘ squared.

Substituting in our values, we have 8.04 is equal to zero multiplied by four plus a half multiplied by π multiplied by four squared. The right-hand side simplifies to eight π. We can then divide through by eight, giving us π 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 πΉ minus three-quarters π
is equal to 12.06.

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

Distributing the parentheses, we have four-fifths πΉ minus 88.2 plus nine twentieths πΉ 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 πΉ plus nine twentieths πΉ is equal to five-quarters or 1.25πΉ. This is equal to 100.26. We can then divide through by 1.25 such that πΉ is equal to 80.208.

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