A uniform rod 2m long has a force 80N applied downward at one end. Another force 6N applied downward at the other end. At what point under the rod should the pivot be placed to keep it in equillibrim?
To find the point under the rod where the pivot should be placed to keep it in equilibrium, we need to consider the torque produced by the 80N force and the 6N force.
Torque is given by the equation:
Torque = Force x Distance
Since the rod is uniform, we can assume its weight acts at its center. The weight of the rod can be considered as a single force of 40N acting downward at its midpoint (halfway between the two ends of the rod).
Let's denote the distance from one end of the rod to the pivot as "x" and the distance from the other end to the pivot as "2 - x" (since the rod is 2m long).
The torque produced by the 80N force is:
Torque1 = 80N x (2 - x)
The torque produced by the 6N force is:
Torque2 = 6N x x
For the rod to be in equilibrium, the total torque must be zero. So we can set up the equation:
Torque1 + Torque2 = 0
80N x (2 - x) + 6N x x = 0
Simplifying the equation:
160N - 80N x + 6N x^2 = 0
Since this is a quadratic equation, we need to solve for x. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6N, b = -80N, and c = 160N.
Plugging in the values and solving the equation, we find:
x = (80 ± √((-80)^2 - 4 * 6 * 160)) / (2 * 6)
Simplifying further:
x = (80 ± √(6400 - 3840)) / 12
x = (80 ± √2560) / 12
x = (80 ± 16√10) / 12
Simplifying even more:
x = 20 ± 4√10 / 3
Therefore, the pivot should be placed at a distance of (20 + 4√10) / 3 or (20 - 4√10) / 3 from one end of the rod to keep it in equilibrium.