Question Video: Analysis of the Equilibrium of a Beam Resting between a Rough Wall and a Rough Ground Mathematics

Video Transcript

A uniform beam 𝐴𝐵 of weight 106 newtons is resting with its end 𝐴 on a rough horizontal ground and with its end 𝐵 against a rough vertical wall, where the coefficient of friction between the beam and the wall is four times that between the beam and the ground. If the beam is about to move when it is inclined to the wall at an angle whose tan is 14 over 45, determine the reaction of the wall rounded to two decimal places.

Let’s begin with a diagram of the scenario. We have the uniform beam 𝐴𝐵 with its end 𝐴 resting on the horizontal ground and its end 𝐵 resting against the wall. The angle between the beam and the wall is 𝜃 whose tan is 14 over 45. Therefore, 𝜃 is equal to the inverse tan of 14 over 45. The beam has a weight of 106 newtons, which acts straight downwards from the beam’s center of mass, which in this case, since the beam is uniform, is from the very center of the line 𝐴𝐵. The coefficient of friction between the beam and the ground is unknown. Let’s call this 𝜇. And the coefficient of friction between the beam and the wall is four times this, so this is four 𝜇.

The question asks us to determine the reaction of the wall. This is the force 𝑅 w which acts on the end 𝐵 of the beam directly out from the wall and horizontally. The beam is given to be at rest. So we can balance all of the horizontal and vertical components of all of the forces acting on the beam.

First, we need to determine what all of these forces are. In addition to the reaction force from the wall, there will also be a reaction force from the ground, 𝑅 G, which acts directly away from the ground vertically upwards. There is a frictional force 𝐹 w acting on the end 𝐵 of the beam. This opposes the direction of motion of the beam along the wall. Since the beam is about to slip, the direction of motion would be directly down the wall. Therefore, 𝐹 w acts directly up the wall. Likewise, there is another frictional force 𝐹 G acting on the beam at the point 𝐴. Again, this frictional force opposes the direction of motion of the beam. And since the beam is about to slip, its motion would be in the rightward direction along the ground. So 𝐹 G acts in the leftward direction.

All of these forces are either directly horizontal or directly vertical. Therefore, they are perpendicular or parallel to each other. And we can resolve their horizontal and vertical components separately. Starting with the horizontal forces then, to the left, we have the frictional force of the ground, 𝐹 G. And to the right, we have the reaction force of the wall, 𝑅 w. So 𝐹 G is equal to 𝑅 w. And for the vertical forces, in the upward direction, we have the frictional force of the wall 𝐹 w and the reaction force of the ground 𝑅 G. And in the downward direction, we have the weight of the beam, 106 newtons. So 𝐹 w plus 𝑅 G equals 106.

Now recall that the coefficient of friction is defined by the equation 𝐹 equals 𝜇𝑅, where the object is at the point of slip, 𝑅 is the reaction force on the object, and 𝐹 is the frictional force. In this scenario then, 𝐹 G is equal to the coefficient of friction of the ground, 𝜇, multiplied by the reaction force of the ground, 𝑅 G. And 𝐹 w is equal to the coefficient of friction of the wall, four 𝜇, multiplied by the reaction force of the wall, 𝑅 w. We can therefore rewrite both of these equations as 𝜇𝑅 G equals 𝑅 w and four 𝜇𝑅 w plus 𝑅 G equals 106. But we have a problem. We now have three unknowns in this system: 𝜇, 𝑅 G, and 𝑅 w. But we only have two equations.

To solve a system of equations for any of the variables, we need at least the same number of equations as there are unknowns. We therefore need to find more different equations for each of these unknowns in this scenario. In a system in equilibrium, it is not only the forces that are balanced but also the moments of the forces about any point in space. For instance, the moments on this beam are balanced about the point 𝐴 and the point 𝐵. If we therefore calculate the moments of all of the forces about the points 𝐴 and 𝐵, they should all be balanced as well.

Let’s begin by calculating the moments of the forces about the point 𝐴, defining a positive moment as being in the counterclockwise direction. Recall that the moment of a force 𝑀 is equal to the magnitude of the force 𝐹 multiplied by the perpendicular distance 𝑑 between the line of action of the force and the pivot point 𝑂. The perpendicular distances between the lines of action of 𝑅 G and 𝐹 G will clearly be equal to zero since they are acting from the point 𝐴.

The perpendicular distance between the line of action of the weight and the pivot point 𝐴 will be this distance here, 𝑑 one. This distance will be half the length of the beam, which we will call 𝑙, multiplied by the sine of the angle between the beam and the wall, 𝜃. The perpendicular distance between the line of action of the reaction force of the wall 𝑅 w and the pivot point 𝐴 will be this distance here, 𝑑 two. This distance will be the length of the beam 𝑙 multiplied by the cosine of the angle between the beam and the wall 𝜃. And finally, the perpendicular distance between the line of action of the frictional force of the wall 𝐹 w and the pivot point 𝐴 will be given by this distance here, 𝑑 three. This distance will be equal to the length of the beam 𝑙 multiplied by the sine of the angle between the beam and the wall 𝜃.

Going through all the moments then, starting with the moment of the weight of the beam, we have the magnitude of the force, which is 106 newtons, multiplied by the perpendicular distance, 𝑑 one, equal to 𝑙 over two sin 𝜃. Clearly this moment will be in the anticlockwise direction, so this is a positive moment.

Next, for the moment of the reaction force of the wall 𝑅 w, we have the magnitude 𝑅 w itself multiplied by 𝑑 two, which is equal to 𝑙 cos 𝜃. 𝑅 w will clearly have a clockwise turning force about the point 𝐴. So this is a negative moment. And finally, we have the moment of the frictional force of the wall, 𝐹 w, which is equal to the magnitude, 𝐹 w itself, multiplied by the distance 𝑑 three, which is 𝑙 sin 𝜃. Once again, this will be a clockwise turning force about the point 𝐴. So this is a negative moment. Since the system is at rest, the sum of the moments about any point including the point 𝐴 will be equal to zero.

We can now simplify this equation. We begin by noting that we have a common factor of 𝑙 in all three turns on the left-hand side. And since 𝑙 is nonzero, we can divide by 𝑙 on both sides to cancel it out completely. Since the beam is about to slip, 𝐹 w is also equal to four 𝜇𝑅 w. And finally, in the first term, we can simplify by taking 106 multiplied by the one-half to give 53. So this gives us 53 sin 𝜃 minus 𝑅 w cos 𝜃 minus four 𝜇𝑅 w sin 𝜃 is equal to zero.

But we can simplify this further to make this even better. 𝜃 is clearly not equal to 90 degrees since its tangent is well defined. Therefore, we can divide both sides by cos 𝜃 since cos 𝜃 will be nonzero. The cos 𝜃 will cancel with the second term. And for the other two terms, sin 𝜃 over cos 𝜃 is identically equal to tan 𝜃. And we are given in the question that tan 𝜃 is equal to 14 over 45. This gives us 53 multiplied by 14 over 45, which is equal to 742 over 45, minus 𝑅 w minus four times 14 over 45, which is equal to negative 56 over 45, multiplied by 𝜇𝑅 w is equal to zero. We can simplify this one more time by multiplying by 45 to give us 742 minus 45𝑅 w minus 56𝜇𝑅 w is equal to zero. We could also factorize the 𝑅 w, but it will be more useful to keep these terms separate in just a moment.

We now have three equations for the three unknowns, which can be solved to give values for each of them. We can go about solving these equations by eliminating the variables in any order that we choose. It turns out that solving for the coefficient of friction 𝜇 is slightly easier than other ways. So let’s try this way.

If we eliminate 𝑅 G from the second and third equation, we will then be able to solve for both 𝜇 and what we want, 𝑅 w. So we begin by rearranging the first equation to give 𝑅 G equals 𝑅 w over 𝜇. We can substitute this into the second equation here, where 𝑅 G appears. And then we have two equations in 𝜇 and 𝑅 w. So for the second equation, we now have four 𝜇𝑅 w plus 𝑅 w over 𝜇 equals 106. We can rearrange the second equation by multiplying by 𝜇 on both sides, which gives four 𝜇 squared 𝑅 w plus 𝑅 w equals 106𝜇. We have a common term of 𝑅 w on the left-hand side. So we can factorize this to give 𝑅 w multiplied by four 𝜇 squared plus one is equal to 106𝜇. Dividing both sides by four 𝜇 squared plus one gives us 𝑅 w equals 106𝜇 divided by four 𝜇 squared plus one.

We now have an explicit expression for 𝑅 w in terms of the coefficient of friction 𝜇. So we will be able to come back to this later after solving for 𝜇 to calculate 𝑅 w easily. For now, we can substitute this expression for 𝑅 w into the final equation, which will give us an equation only in 𝜇, which can be solved. So this gives us 742 minus 45 times 106𝜇 over four 𝜇 squared plus one minus 56𝜇 times 106𝜇 over four 𝜇 squared plus one is equal to zero.

Now, we can simplify this somewhat if we like by noting that 742 is 106 times seven. Therefore, we have a common factor of 106 in all three terms. Dividing through by 106 will give us seven for the first term, will cancel out the 106 in the second and third terms. This gives us seven minus 45𝜇 over four 𝜇 squared plus one minus 56𝜇 squared over four 𝜇 squared plus one is equal to zero. We can multiply both sides by four 𝜇 squared plus one, which will cancel out the denominator on the second and third terms. This gives us seven times four 𝜇 squared plus one minus 45𝜇 minus 56𝜇 squared is equal to zero.

Expanding the parentheses and collecting like terms in 𝜇 gives us negative 28𝜇 squared minus 45𝜇 plus seven is equal to zero. This is a quadratic in 𝜇, which can be solved using the quadratic formula. The quadratic formula is given by 𝜇 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎, where 𝑎 is the coefficient of the 𝜇 squared term, so in this case negative 28. 𝑏 is the coefficient of the 𝜇-term, so negative 45. And 𝑐 is the constant coefficient, which is seven. So this gives us 𝜇 equals 45 plus or minus the square root of negative 45 squared minus four times negative 28 times seven all over two times negative 28.

The term under the square root comes to 2809, and the denominator comes to negative 56. The square root of 2809 is plus or minus 53. The coefficient of friction 𝜇 must be greater than or equal to zero. So clearly, we need to make the numerator negative so that 𝜇 is overall positive. Therefore, we will be taking the negative square root. This gives us negative eight over negative 56, so 𝜇 is equal to one-seventh. We can now substitute this value of 𝜇 into our equation for 𝑅 w, and we will have our answer. So we have 𝑅 w equals 106 multiplied by one over seven all divided by four times one over seven squared plus one. Performing this calculation gives us 14 exactly, so we have our final answer. The reaction of the wall rounded to two decimal places, 𝑅 w, equals 14.00 and the unit is newtons.