A horizontal plank 4.12 m long and having mass 26.8 kg rests on two pivots, one at the left end and a second 1.03 m from the right end. Find the magnitude of the force exerted on the plank by the second pivot.
N
To find the magnitude of the force exerted on the plank by the second pivot, we can use the principle of rotational equilibrium.
The principle of rotational equilibrium states that the sum of the torque (rotational force) acting on an object must be zero. In this case, the torques created by the weight of the plank and the force exerted by the first pivot are balanced by the torque created by the force exerted by the second pivot.
First, let's calculate the torque created by the weight of the plank. The weight of an object is given by the formula:
Weight = mass * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
Weight of the plank = mass of the plank * acceleration due to gravity
= 26.8 kg * 9.8 m/s^2
= 262.64 N
The torque created by the weight of the plank is given by:
Torque = weight of the plank * distance from the pivot to the center of mass
The distance from the left end to the center of mass is half the length of the plank:
Distance from the left end to the center of mass = length of the plank / 2
= 4.12 m / 2
= 2.06 m
Therefore, the torque created by the weight of the plank is:
Torque = 262.64 N * 2.06 m
= 540.2344 N·m
Since the plank is in rotational equilibrium, the torque created by the force exerted by the first pivot must be equal to the torque created by the weight of the plank.
Now, let's calculate the torque created by the force exerted by the first pivot. Since the first pivot is at the left end, the torque created by it is zero.
Since the torque created by the force exerted by the first pivot is zero, the torque created by the second pivot must be equal to the torque created by the weight of the plank.
The distance from the right end to the second pivot is given as 1.03 m.
Now, let F be the magnitude of the force exerted by the second pivot. The torque created by the second pivot is given by:
Torque = F * distance from the pivot to the center of mass
The distance from the right end to the center of mass is the total length of the plank minus the distance from the right end to the second pivot:
Distance from the right end to the center of mass = length of the plank - distance from the right end to the second pivot
= 4.12 m - 1.03 m
= 3.09 m
Setting the torque created by the second pivot equal to the torque created by the weight of the plank gives us:
F * 3.09 m = 540.2344 N·m
Now, we can solve for F by dividing both sides of the equation by 3.09 m:
F = 540.2344 N·m / 3.09 m
= 174.9476 N
Therefore, the magnitude of the force exerted on the plank by the second pivot is approximately 174.95 N.