Question Video: Studying the Equilibrium of a Horizontal Rod Resting by the Means of a Support and a String Mathematics

A uniform rod 𝐴𝐡 is resting horizontally at its ends on two supports. And it carries a weight of 17 N at a point that is 96 cm away from 𝐡. If a weight of 64 N is suspended at a point on the rod that’s π‘₯ cm away from 𝐡, the reaction at 𝐡 is double that at 𝐴. Given that the rod is 144 cm long and weighs 30 N, determine the value of π‘₯ and the magnitude of the reaction 𝑅_(𝐴) at 𝐴.

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Video Transcript

A uniform rod 𝐴𝐡 is resting horizontally at its ends on two supports. And it carries a weight of 17 newtons at a point that is 96 centimeters away from 𝐡. If a weight of 64 newtons is suspended at a point on the rod that’s π‘₯ centimeters away from 𝐡, the reaction at 𝐡 is double that at 𝐴. Given that the rod is 144 centimeters long and weighs 30 newtons, determine the value of π‘₯ and the magnitude of the reaction 𝑅 sub 𝐴 at 𝐴.

Let’s begin by drawing a diagram to model the scenario. We are told that the rod rests horizontally on two supports at its ends 𝐴 and 𝐡. The rod is 144 centimeters long. It weighs 30 newtons. Since the rod is uniform, its weight acts at the midpoint. This means there is a 30-newton downward force 72 centimeters from 𝐴 and 72 centimeters from 𝐡. We are also told that there is a 17-newton weight 96 centimeters away from 𝐡. And finally a weight of 64 newtons is suspended at a point that is π‘₯ centimeters from 𝐡.

There will be reaction forces acting vertically upwards at 𝐴 and 𝐡. We will call these 𝑅 sub 𝐴 and 𝑅 sub 𝐡. We are told that the reaction at 𝐡 is double that at 𝐴. Therefore, 𝑅 sub 𝐡 is equal to two 𝑅 sub 𝐴. It is this reaction force 𝑅 sub 𝐴 together with the value π‘₯ we are trying to calculate.

Since the rod is resting, we can assume it is in equilibrium. And this means that the sum of the forces acting on it equals zero. The sum of the moments also equals zero, where the moment of the force is equal to 𝐹 multiplied by 𝑑. 𝐹 is the force acting at a given point, and 𝑑 is the perpendicular distance from that point to the point at which we are taking moments.

Let’s begin by resolving vertically, where the positive direction is vertically upwards. The forces 𝑅 sub 𝐴 and 𝑅 sub 𝐡 act in the positive direction, whereas the 17-newton, 30-newton, and 64-newton weights all act in the negative direction. As these forces sum to zero, we have 𝑅 sub 𝐴 plus 𝑅 sub 𝐡 minus 17 minus 30 minus 64 equals zero. We know that 𝑅 sub 𝐡 is equal to two 𝑅 sub 𝐴. So our equation simplifies to 𝑅 sub 𝐴 plus two 𝑅 sub 𝐴 minus 111 is equal to zero. Adding 111 to both sides, we have three 𝑅 sub 𝐴 is equal to 111. We can then divide through by three such that 𝑅 sub 𝐴 is equal to 37 newtons. The magnitude of the reaction at point 𝐴 is 37 newtons.

We will now take moments about a point on the rod. Whilst we could do this at any point, we will do so at point 𝐡, as the distance π‘₯ is being measured from point 𝐡. We will take moments in the counterclockwise direction to be positive. The moment of the reaction force of 𝐡 will be zero, as this is acting at point 𝐡. And the distance will therefore be equal to zero.

The 64-newton force acts in a counterclockwise direction about point 𝐡 and will therefore have a positive moment. This is equal to 64 multiplied by π‘₯. The 30-newton force also acts in the counterclockwise direction. This will have a moment of 30 multiplied by 72, as it acts a distance of 72 centimeters from 𝐡. The 17-newton force will have a positive moment equal to 17 multiplied by 96. Finally, the reaction force at 𝐴 acts in a clockwise direction about point 𝐡 and will therefore have a negative moment. This is equal to negative 𝑅 sub 𝐴 multiplied by 144.

We have already calculated that 𝑅 sub 𝐴 is equal to 37 newtons. And we know that the sum of all the moments equals zero. Our equation simplifies to 64π‘₯ plus 2160 plus 1632 minus 5328 equals zero. This is equal to 64π‘₯ minus 1536 equals zero. We can then add 1536 to both sides. And then dividing through by 64, we have π‘₯ is equal to 24. The weight of 64 newtons is suspended at a point on the rod that is 24 centimeters away from 𝐡. Our two answers are π‘₯ equals 24 and 𝑅 sub 𝐴 equals 37 newtons.

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