# Question Video: An Inclined Force Acting on a Body in Equilibrium on a Horizontal Rough Plane Mathematics

A body of weight 45 N rests on a rough horizontal plane. If a horizontal force of 11 N acted on the body, it would be on the point of moving. Instead, a force, whose line of action was inclined to the horizontal at an angle of 60°, was acting on the body. Given that the body was on the point of moving, find the magnitude of this force rounding your answer to two decimal places if required.

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### Video Transcript

A body of weight 45 newtons rests on a rough horizontal plane. If a horizontal force of 11 newtons acted on the body, it would be on the point of moving. Instead, a force whose line of action was inclined to the horizontal at an angle of 60 degrees was acting on the body. Given that the body was on the point of moving, find the magnitude of this force, rounding your answer to two decimal places if required.

There are two scenarios we need to consider in this question. In both cases, the body is on the point of moving and is therefore in equilibrium. This means that when we resolve vertically and horizontally, the sum of our forces will equal zero. We also know that the maximum frictional force 𝐹 𝑟 between a body and a rough surface is equal to 𝜇 multiplied by 𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the normal reaction force.

Let’s begin by sketching a free-body diagram of the first scenario. We are told that the body weighs 45 newtons. Therefore, this force will act vertically downwards. By considering Newton’s third law, there will be a normal reaction force acting vertically upwards. An 11-newton horizontal force acts on the body. And the frictional force 𝐹 𝑟 will act in the opposite direction to this. Resolving vertically, we have 𝑅 minus 45 is equal to zero, where the positive direction is vertically upwards. Adding 45 to both sides of this equation, we have 𝑅 equals 45. The normal reaction force acting on the body is 45 newtons.

Resolving horizontally, we have 11 minus 𝐹 𝑟 is equal to zero. Adding the frictional force to both sides of this equation, we have 𝐹 𝑟 equals 11. The frictional force is equal to 11 newtons. We can use these values to calculate the coefficient of friction 𝜇. We have the equation 11 is equal to 𝜇 multiplied by 45. And dividing both sides of the equation by 45 gives us 𝜇 is equal to 11 over 45.

Let’s now consider the second scenario when, instead of an 11-newton force, we have a force inclined to the horizontal at an angle of 60 degrees. This force 𝐹 will have both a horizontal and vertical component. And we can calculate these using our knowledge of right-angled trigonometry. The horizontal component 𝑥 is adjacent to the 60-degree angle, whereas the vertical component 𝑦 is opposite the 60-degree angle. We know that the cos of any angle 𝜃 in a right triangle is equal to the adjacent over the hypotenuse. This means that the cos of 60 degrees is equal to 𝑥 over 𝐹.

The cos of 60 degrees is equal to one-half. And multiplying through by 𝐹, we have 𝑥 is equal to a half 𝐹. The horizontal component of our force is one-half 𝐹. In the same way, since the sin of angle 𝜃 is equal to the opposite over the hypotenuse, we have the sin of 60 degrees is equal to 𝑦 over 𝐹. The sin of 60 degrees is root three over two. Therefore, 𝑦 is equal to root three over two 𝐹. This is the vertical component of our force.

We can now resolve vertically and horizontally once again. As the body is still in equilibrium, resolving vertically, we have 𝑅 plus root three over two 𝐹 minus 45 is equal to zero. Resolving horizontally, we have a half 𝐹 minus the frictional force 𝐹 𝑟 equals zero. Since this frictional force is equal to 𝜇 multiplied by 𝑅 and 𝜇 is equal to 11 over 45, this can be rewritten as a half 𝐹 minus 11 over 45 𝑅 is equal to zero.

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. And we begin by making 𝑅 the subject of the first equation. We add 45 and subtract root three over two 𝐹 from both sides such that 𝑅 is equal to 45 minus root three over two 𝐹. We can then substitute this expression for 𝑅 into the second equation. This gives us a half 𝐹 minus 11 over 45 multiplied by 45 minus root three over two 𝐹 is equal to zero. We can then distribute the parentheses, giving us a half 𝐹 minus 11 plus 11 root three over 90 𝐹 is equal to zero.

We will now clear some space and solve this equation. Firstly, we can add 11 to both sides. We can then add the coefficients of 𝐹 such that 45 plus 11 root three over 90 𝐹 is equal to 11. We can then divide through by 45 plus 11 root three over 90, giving us 𝐹 is equal to 15.4560 and so on. We are asked to round our answer to two decimal places. This means that 𝐹 is approximately equal to 15.46. The magnitude of the force 𝐹 is 15.46 newtons.