Question Video: Finding the Coefficient of Friction between a Body in Equilibrium and a Rough Horizontal Plane | Nagwa Question Video: Finding the Coefficient of Friction between a Body in Equilibrium and a Rough Horizontal Plane | Nagwa

# Question Video: Finding the Coefficient of Friction between a Body in Equilibrium and a Rough Horizontal Plane Mathematics • Third Year of Secondary School

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A body weighing 8 N rests on a rough horizontal plane. A force of 6β2 N is acting on the body such that its line of action makes an angle of 45Β° downward to the horizontal. Given that, as a result, the body is on the point of moving, find the coefficient of friction π between the body and the plane and the angle of friction π stating your answer to the nearest minute.

08:15

### Video Transcript

A body weighing eight newtons rests on a rough horizontal plane. A force of six root two newtons is acting on the body such that its line of action makes an angle of 45 degrees downward to the horizontal. Given that, as a result, the body is on the point of moving, find the coefficient of friction π between the body and the plane and the angle of friction π, stating your answer to the nearest minute.

Alright, so in this scenario, we have a body at rest on a rough horizontal surface, and weβre told that thereβs a force of six root two newtons acting on the body at an angle 45 degrees below the horizontal. Based on this, the two quantities we want to solve for are the coefficient of friction π between the body and the surface and this angle called π, which is the angle of friction. Weβll get to this angle later on. First, letβs focus on the coefficient of friction π.

To understand this coefficient, weβll need to know the forces, all of them, that are acting on our body. Along with the applied force of six root two newtons, thereβs a weight force acting straight downward, a reaction force acting upward, where here the length of our arrows are not drawn to scale, and, finally, thereβs a frictional force that opposes the horizontal direction of our applied six root two newton force.

Marking down that our weight force, weβll call it π, is eight newtons, we can clear some space to work and begin our analysis by noting that this body is at rest. This means that the sum of horizontal forces acting on the body, where weβre saying that forces to the right are positive, is zero. And likewise, the sum of all vertical forces acting on it are zero too. We can therefore write out an equation for all of the horizontal forces involved. In what weβve called the positive direction, thereβs the horizontal component of our six root two newton force. Leaving off units, that force equals six times the square root of two times the cos of 45 degrees.

Then, from this we subtract the frictional force πΉ, which is entirely horizontal. And because our body is at rest, the sum of these forces is zero. This implies that πΉ equals six root two times the cos of 45 degrees. And πΉ, we recall, is the frictional force which in general is equal to a coefficient of friction π multiplied by the reaction force acting on a given body. We can therefore substitute π times π in for πΉ in this equation. And if we then divide both sides by the reaction force π, canceling that factor on the left, weβve now got an expression where π is the subject.

To evaluate this expression, weβll need to know what the reaction force π is on our body. We can see what that is by considering the vertical forces that act on it. Just like we did for the horizontal forces, we can write a vertical force balance equation. We have our reaction force π in the positive or upward direction. From that, we subtract our weight force π, and weβll be careful not to leave out the vertical component of our six root two newton force. Leaving out the units, thatβs six root two times the sin of 45 degrees. And as we said, because our body is an equilibrium, all these forces add to zero.

Considering this expression, we can see that if we add π and six root two times the sin of 45 degrees to both sides, then weβll see that π equals π plus six root two sin 45 degrees. And so we can take the right-hand side of this expression and substitute it in for π. The coefficient of friction is given by this expression then. And if we clear some space to evaluate it, we can notice two things about the right-hand side. First, the weight force π equals eight newtons and, second, that the cos and the sin of 45 degrees equals the square root of two over two.

Leaving out units and making these substitutions, notice how in our numerator we have the square root of two times the square root of two then divided by two, that all comes out to one. And the same thing happens in the second term of our denominator. This whole fraction then simplifies to six over eight plus six, which is six over 14 or three over seven. This, then, is our value for the coefficient of friction π.

Our next step is to solve for this quantity called the angle of friction. If we go back to a simplified sketch of our body at rest, we can imagine very gradually inclining this surface. As we do this, we know the body at rest will be approaching the point where it starts to slide downhill. Eventually, our angle of inclination gets big enough so that thatβs exactly what happens. Right before our body begins to slide, weβre at an angle of inclination called the angle of friction. This is where the force of static friction acting up the incline exactly balances out the component of the weight force that acts down the slope.

As our problem statement said, itβs this angle π that we want to solve for. To start doing that, letβs set up the convention that the positive π₯-direction in this scenario is down the slope and the positive π¦-direction is perpendicularly away from it. Using this framework, we can look at forces in the π₯- and π¦-directions separately. Speaking of those forces, letβs make sure weβve included all the forces acting on our body. In addition to the weight force and the frictional force, thereβs a reaction or normal force weβll call π that acts on it as well. True to its name, this force is normal or perpendicular to the slope of the incline. Therefore, itβs entirely in the positive π¦-direction.

Because our slope is inclined at the angle of friction, we know that our body is in equilibrium. This tells us that the net force acting on our body is zero. When it comes to forces in the π₯-direction, we can see that there are two. First, thereβs the π₯-component of our weight force drawn here in orange. And then thereβs the frictional force, which acts in the negative π₯-direction. In order to solve for the π₯-component of the weight force, weβll need to recognize that this interior angle in what is a right triangle is equal to the angle of inclination of our plane in the first place. Itβs π. This, by the way, is always the case, even when our angle is not the angle of friction.

Since the weight force π is the length of the hypotenuse of this right triangle, the length of the triangle side opposite the angle π is equal to π times the sin of π. From this force, we subtract the force of friction. And again, because our body is in equilibrium, these forces sum to zero. We could say then that our frictional force πΉ equals π times the sin of π and that frictional force in general equals π times the reaction force π. We can make this substitution then into our equation. Now itβs π that we want to solve for. In terms of the other variables in this expression, we know the weight force π and weβve solved for the coefficient of friction π. But we donβt yet know the reaction force π for this inclined scenario. We can solve for it, though, by considering the forces that act on our body in the π¦-direction.

Here we have the reaction force π which is entirely in the positive π¦-direction. And then we have the π¦-component of the weight force here. Just as the π₯-component of that force was π times the sin of π, when we calculate the π¦-component, thatβs π times the cos of that angle. Adding these two forces together gives us zero because our body is in equilibrium, so we can say that π equals π times the cos of π. We can now make this substitution for π. Doing that gives us π times π times the cos of π equals π times the sin of π. Note that the weight force is common to both sides and therefore cancels out.

Next, we divide both sides by the cos of π, canceling that factor on the left. And then notice what we have on the right, the sin of an angle divided by the cos of that same angle. In general, this equals the tan of that angle. We can write then that π, the coefficient of friction, equals the tan of π, the angle of friction. This implies that π equals the inverse tan of π. And now letβs recall that π is three-sevenths. When we calculate the inverse tan of three-sevenths and then round the answer we get, which is an angle to the nearest minute, we find a result of 23 degrees and 12 minutes. So the coefficient of friction between our body and the surface it rests on is three-sevenths, and the angle of friction, the maximum angle at which it can be inclined before it slides, is 23 degrees and 12 minutes.

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