Video: Finding the Distance at Which a Weight Is Attached to a Rod Resting Horizontally on Two Supports

A uniform rod 𝐴𝐡 having a weight of 64 N and a length of 168 cm is resting horizontally on two identical supports at its ends. A weight of magnitude 56 N is suspended at a point on the rod that is π‘₯ cm away from 𝐴. If the magnitude of the reaction at 𝐴 is double that at 𝐡, determine the value of π‘₯.

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Video Transcript

A uniform rod 𝐴𝐡 having a weight of 64 newtons and a length of 168 centimetres is resting horizontally on two identical supports at its ends. A weight of magnitude 56 newtons is suspended at a point that is π‘₯ centimetres away from 𝐴. If the magnitude of the reaction at 𝐴 is double that at 𝐡, determine the value of π‘₯.

In order to solve this problem, we will resolve vertically and take moments about a point. We begin by drawing a diagram of the rods. There are two supports at either end of the rod. These will have a reaction force 𝑅 𝐴 and 𝑅𝐡. We’re told that the reaction at 𝐴 is double the reaction at 𝐡. If we call the reaction at 𝐡 𝑅, then the reaction at 𝐴 will be two 𝑅. We’re told that the rod is uniform and has a weight of 64 newtons. This means that the force of 64 newtons will act at the centre of the rod. The rod is 168 centimetres in length. This means that the 64-newton force will be 84 centimetres from 𝐴, as one-half of 168 is 84.

There is a weight of magnitude 56 newtons suspended from the rod. We’re told that this is π‘₯ centimetres from 𝐴. We can now begin to solve the problem by resolving vertically. The forces going upwards will be equal to the forces going downwards. So two 𝑅 plus 𝑅 is equal to 56 plus 64. The left-hand side simplifies to three 𝑅, and 56 plus 64 equals 120. Dividing both sides by three gives us 𝑅 is equal to 40. This means that the reaction force at 𝐡 is 40 newtons and the reaction force at 𝐴 is 80 newtons, as two multiplied by 40 is equal to 80.

We will now take moments about a point. In this question, we will choose point 𝐴. We recall that the moment of a force is equal to the force multiplied by the distance from some point. We also recall that when the rod is in equilibrium, the clockwise moments equal the anticlockwise moments. In this question, the 56-newton force and the 64-newton force are moving clockwise around point 𝐴, whereas the 40-newton force is moving anticlockwise. The sum of the moments in the clockwise direction is, therefore, 56 multiplied by π‘₯ plus 64 multiplied by 84. The anticlockwise moments are 40 multiplied by 168. 64 multiplied by 84 is 5376, and 40 multiplied by 168 is 6720. Bouncing this equation gives us 56π‘₯ is equal to 1344. Our final step is to divide both sides by 56. 1344 divided by 56 is equal to 24. The value of π‘₯ is, therefore, equal to 24 centimetres.

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