Question Video: Analysis of the Equilibrium of a Uniform Rod Free to Rotate with a Nail on a Point With a Couple Acting on Its Endpoints Mathematics

𝐴𝐡 is a uniform rod with length 6 cm. It is free to rotate about a smooth nail in a small hole in the rod at a point 𝐢 between 𝐴 and 𝐡, where 𝐴𝐢 = 2 cm. The rod is in equilibrium, laying horizontally, under the action of two forces, each of magnitude 8 N, acting at either end at an angle of 30Β° with the rod as shown in the figure below. Find the weight of the rod π‘Š and the magnitude of the reaction of the nail 𝑅.

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Video Transcript

𝐴𝐡 is a uniform rod with length six centimeters. It is free to rotate about a smooth nail in a small hole in the rod at a point 𝐢 between 𝐴 and 𝐡, where 𝐴𝐢 is equal to two centimeters. The rod is in equilibrium, laying horizontally, under the action of two forces, each of magnitude eight newtons, acting at either end at an angle of 30 degrees with the rod as shown in the figure below. Find the weight of the rod π‘Š and the magnitude of the reaction of the nail 𝑅.

As the rod is in equilibrium, we know that the sum of the forces in a horizontal π‘₯-direction are equal to zero and the sum of the net forces in the vertical 𝑦-direction are also equal to zero. The sum of the moments about any point on the rod will also equal zero. We will take the positive direction to be to the right horizontally, up vertically, and counterclockwise when taking moments about a point. We have a reaction force 𝑅 acting vertically upwards at point 𝐢. As the rod is uniform, its weight will act vertically downwards at the center of the rod. We will call this point π‘š as it is the midpoint of the rod and it’s three centimeters from 𝐴.

The two forces acting at 𝐴 and 𝐡 are a coplanar couple. This means that they’re equal in magnitude, in this case, eight newtons, but act in the opposite direction. This means that when resolving vertically and horizontally, the two forces will cancel. We will now clear some space so we can set up our equations. We will begin by resolving vertically. In order to do so, we need to calculate the vertical component of the two eight-newton forces. We will do this using our knowledge of right-angle trigonometry. We know that sin of angle πœƒ is equal to the opposite over the hypotenuse.

The vertical force at point 𝐴 labeled π‘₯ satisfies the equation sin of 30 degrees is equal to π‘₯ over eight. We know that the sin of 30 degrees is equal to one-half. Multiplying both sides of this equation by eight gives us π‘₯ is equal to four. The downward vertical force at point 𝐴 is equal to four newtons. We have the same equation at point 𝐡. This time, the sin of 30 degrees is equal to 𝑦 over eight. 𝑦 is also equal to four newtons. There is a force acting vertically upwards at point 𝐡 equal to four newtons. We now have four forces acting in the vertical direction. As the sum of these forces must equal zero and the positive direction is upwards, we have negative four plus 𝑅 minus π‘Š plus four is equal to zero.

As previously mentioned, the coplanar couple forces will cancel. Adding π‘Š to both sides, we see that 𝑅 is equal to π‘Š. The reaction force at 𝑅 and the weight force at π‘š will be another coplanar couple as they will be equal in magnitude and are opposite in direction. We will now take moments about a point on the rod. And in this case, we will choose point 𝐴. We know that the sum of the moments will be equal to zero and the moments acting in a counterclockwise direction will be positive. The moment of each force is equal to the force multiplied by the perpendicular distance. We begin with the force 𝑅. The moment here is equal to 𝑅 multiplied by two centimeters.

Next, we have the weight force. The moment here is equal to π‘Š multiplied by three. This is negative as it is moving in a clockwise direction. Finally, we have the four-newton force multiplied by six centimeters in the positive direction. The sum of these three moments is equal to zero. This means the equation simplifies to two 𝑅 minus three π‘Š plus 24 is equal to zero. As 𝑅 is equal to π‘Š, this can be rewritten as two π‘Š minus three π‘Š plus 24 equals zero. Collecting like terms gives us negative π‘Š plus 24 is equal to zero.

Finally, adding π‘Š to both sides gives us π‘Š is equal to 24. The weight of the rod is equal to 24 newtons, and the magnitude of the reaction of the nail 𝑅 is also 24 newtons.

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