Video Transcript
𝐴𝐵 is a rod having a length of 50
centimeters and a negligible weight. Two coplanar pairs of forces are
acting on the rod as shown in the figure. The first couple consists of two
forces acting perpendicularly to the rod, each of magnitude two kilogram-weight. And the second couple consists of
two forces, each of magnitude 𝐹. Determine the value of 𝐹 that
makes the rod in equilibrium.
We recall that a coplanar couple is
two forces that are equal in magnitude and opposite in direction. In this question, we have two such
couples. In order to calculate 𝐹, we’ll
need to take moments about a point on the rod. We know that when the rod is in
equilibrium, the sum of our moments is equal to zero. We know that a moment is equal to
the force multiplied by the perpendicular distance. This means that we will firstly
need to calculate the vertical components of the forces 𝐹.
Using our knowledge of right-angle
trigonometry, we know that the sine of angle 𝜃 is equal to the opposite over the
hypotenuse. This means that, in this question,
the sin of 45 degrees is equal to 𝑦 over 𝐹. The vertical components are
therefore equal to 𝐹 multiplied by sin of 45 degrees. The sin of 45 degrees is equal to
root two over two. We will now take moments about
point 𝐵, taking the counterclockwise direction to be positive.
The first force of magnitude two
newtons will have a moment equal to two multiplied by 10 as it is 10 centimeters
from 𝐵. Next, we have root two over two 𝐹
multiplied by 30. Our final force is acting in a
clockwise direction, giving us negative two multiplied by 50. The sum of these forces equals
zero. Our equation simplifies to 20 plus
15 root two 𝐹 minus 100 equals zero. 20 minus 100 is equal to negative
80, and we can add this to both sides. Dividing both sides by 15 root two
gives us 𝐹 is equal to 16 over three root two. We can then rationalize the
denominator, giving us a value of 𝐹 equal to eight root two over three
kilogram-weight.
