Q.D-1. A uniform metal stick having mass 400 g is suspended from the fixed supports through two vertical light strings of equal lengths fixed at the ends. A small object of mass 100 g is put on the stick at a distance of 60 cm from the left end. Calculate the tensions in the two strings. (g = 10 )
Dear Student ,
Let the tension in the left string T1 and the right string T2
these two tensions must support the total weight of the system, so we have
T1+T2=0.5 kg *9.81m/s/s=4.905 N
the stick must also be in rotational equilibrium, so we have the condition that the sum of torques around any point on the stick must be zero
we can choose any point about which to do our torque analysis, let's choose the left end of the stick since the torque due to T1 will be zero
summing torques around the left end:
the 100g mass exerts a torque of weight x distance = 0.1kgx9.81m/s/s x 0.6m = 0.5886 Nm and this is a clockwise torque
since the stick is uniform, its weight acts as if it is at the center of mass, or at the 0.5m mark; the torque due to the stick is 0.4kgx9.81m/s/sx0.5m=1.962 Nm and this is also a clockwise torque
the torque due to T2 is T2 x 1m and this is counterclockwise, so this torque must balance the other two, and we have:
T2*1m=1.962Nm+0.5886Nm=2.5506N
since T1+T2=4.905N, T2=2.5506N and T1=2.354N
Regards
Let the tension in the left string T1 and the right string T2
these two tensions must support the total weight of the system, so we have
T1+T2=0.5 kg *9.81m/s/s=4.905 N
the stick must also be in rotational equilibrium, so we have the condition that the sum of torques around any point on the stick must be zero
we can choose any point about which to do our torque analysis, let's choose the left end of the stick since the torque due to T1 will be zero
summing torques around the left end:
the 100g mass exerts a torque of weight x distance = 0.1kgx9.81m/s/s x 0.6m = 0.5886 Nm and this is a clockwise torque
since the stick is uniform, its weight acts as if it is at the center of mass, or at the 0.5m mark; the torque due to the stick is 0.4kgx9.81m/s/sx0.5m=1.962 Nm and this is also a clockwise torque
the torque due to T2 is T2 x 1m and this is counterclockwise, so this torque must balance the other two, and we have:
T2*1m=1.962Nm+0.5886Nm=2.5506N
since T1+T2=4.905N, T2=2.5506N and T1=2.354N
Regards