A uniform heavy bar, AB of length 120cm and weighing 5N is supported at one end A and also at C 20cm from B. A load of 10N is suspended at D, which is 20cm form C. Find the force on the support if the system is at equilibrium
Well, if the system is in equilibrium, that means it's not going anywhere. I guess you could say it's really good at "bar-ring" any movement!
To find the force on the support, we need to consider the torques acting on the bar. Since there's no rotation occurring, the sum of all the torques must be zero.
To calculate the torques, we need to consider both the weight of the bar and the weight of the load. The torque due to the weight of the bar can be calculated using the formula:
Torque_bar = weight_bar * distance_AC
Since the bar is horizontal and uniform, its weight can be considered acting at its center of mass, which is at a distance of 60 cm from A.
So, Torque_bar = 5N * 60cm = 300 Ncm
The torque due to the weight of the load can be calculated using a similar formula:
Torque_load = weight_load * distance_CD
In this case, the weight of the load is 10N and the distance from C to D is 20cm.
So, Torque_load = 10N * 20cm = 200 Ncm
Since the system is in equilibrium, the sum of all the torques must be zero.
Torque_bar + Torque_load = 0
300 Ncm + 200 Ncm = 0
The torques are equal in magnitude but opposite in direction, which means they cancel each other out.
So, the force on the support is 0! But don't worry, the support isn't feeling left out – it's just doing its job of keeping everything in balance.