magnitude direction ---Select---no directionclockwisecounter-clockwise Calculate
ID: 2265517 • Letter: M
Question
magnitude direction ---Select---no directionclockwisecounter-clockwise Calculate the total torque (magnitude and direction) for each of the following cases shown below. You can assume the "bar" the masses are attached to is massless. The top point of the triangle represents the pivot point for the system. Be sure to include the magnitude and the direction. Here m = 8 g and d = 16 cm. Here m = 8 g and d = 16 cm. Here m1 = m2 = 8 g and d1 = d2 = 16 cm. Here m1 = m2 = 8 g and d1 = 16 cm and d2 = 1.5 m.Explanation / Answer
a)
T = r x f
Where r is the distance away from the pivot point and f is the force perpendicular to the arm of rotation.
T = r x Fg
T = r x mg
T = (0.12 m)(0.010 kg)(9.81 m/s^2)
T = 0.0118 Nm
You determine direction with the right hand rule. Curl your fingers (of you right hand!) in the direction of the rotation and your thumb will point towards the direction of the torque. So in this case the torque will be into the page.
b)
Exact same thing except the torque is coming out of the page.
c)
Zero torque because the forces are clearly equal and opposite.
d)
T1 = r x (m1)g
T1 = (0.12 m)(0.010 kg)(9.81 m/s^2)
T1 = 0.0118 Nm [into the page]
T2 = r x (m2)g
T2 = (0.7 m)(0.010 kg)(9.81 m/s^2)
T2 = 0.0687 Nm [out of the page]
Let's just define out of the page as positive.
T net = T1 + T2
T net = - 0.0118 Nm + 0.0687 Nm
T net = 0.0569 Nm [out of the page]
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