a meter rod is pivoted at the center 5kg body is attched 10cm from the one end 30cm from the pivot at the other side how much mass be connected to keep the meter in equilibrium
5 kg*40cm = m kg *30 cm
m = 5(4/3) = 20/3 kg
A uniform rod of mass 5kg is an lenghth 50cm is pivoted at the 30cm mark. what force would be applied at the 50cm mark to keep the rod in equilibrium?
To solve this problem, we need to understand the concept of a balanced meter rod.
A balanced meter rod occurs when the sum of the torques acting on the rod is equal to zero. Torque is calculated by multiplying the force applied to an object by the distance from the pivot point. In this case, the pivot point is the center of the meter rod.
Let's denote the mass that needs to be connected to keep the meter rod in equilibrium as "M". Since the 5kg body is attached 10cm from one end and the other mass is 30cm from the pivot, we can represent the torques as follows:
Torque due to the 5kg body = (5kg) * (9.8m/s^2) * (10cm)
Torque due to the mass "M" = (M) * (9.8m/s^2) * (30cm)
Now, since the rod is in equilibrium, the sum of these torques should be equal to zero:
(5kg) * (9.8m/s^2) * (10cm) + (M) * (9.8m/s^2) * (30cm) = 0
To solve for the unknown mass "M", we can rearrange the equation:
(M) * (9.8m/s^2) * (30cm) = - (5kg) * (9.8m/s^2) * (10cm)
Simplifying the equation:
(M) * (30cm) = - (5kg) * (10cm)
Dividing both sides by (30cm):
M = - (5kg) * (10cm) / (30cm)
M = - 5kg / 3
So, to keep the meter rod in equilibrium, a mass of approximately 1.67 kg (or -5/3 kg) should be connected on the other side. Note that the negative sign indicates that the mass should be placed in the opposite direction from the 5kg body to balance the torques.