Question

As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 =  Fp = F =  (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 =  Fp =  F =  (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 =  r5 =

Answer 1

a) Recall, the net torque on the rod must be zero. Thus,

Σt = 0

where

t represents torque

Thus,

t1 + t2 - tr - t3 = 0

t = rF

where

F = force

r = distance

r1F1 + r2F2 - rrFr - r3F3 = 0

r3F3 = r1F1 + r2F2 - rrFr

r3 = (r1F1 + r2F2 - rrFr)/F3

Note,

F1 = T1 = m1g

F2 = T2 = m2g

F3 = T3 = m3g

Thus,

r3 = (r1m1g + r2m2g - rrmrg)/m3g

g cancels out

r3 = (r1m1 + r2m2 - rrmr)/m3

From the information given,

r1 = 10 cm = 10/100 = 0.1 m

r2 = 90 cm = 90/100 = 0.9 m

rr = 100/2 = 50 cm = 50/100 = 0.5 m

m1 = 281 g = 281/1000 = 0.281 kg

m2 = 177g = 0.177 kg

mr = 158g = 0.158 kg

m3 = 200g = 0.2kg

By substituting these values into the equation,

r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2

r3 = 0.542 m

The force exerted by the pin, Fp = mg

g = 9.8

Fp = (m3 - mr - m1 - m2)g

Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8

Fp = - 0.981

Taking the absolute value,

IFpI = 0.981 N

F = - 90 degrees

b) r1F1 + r2F2 - rrFr - r4F4 = 0

r4F4 = r1F1 + r2F2 - rrFr = 0

F4 = (r1F1 + r2F2 - rrFr)/r4

Note,

F1 = T1 = m1g

F2 = T2 = m2g

F3 = T3 = m3g

F4 = T4 = m4g

Thus,

m4g = (r1m1g + r2m2g - rrmrg)/r4

m4g = (r1m1 + r2m2 - rrmr)/r4

r4 = 0.2

By substituting these values into the equation,

m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2

m4 = 0.542 kg

The force exerted by pin is

Fp = (m4 + mr - m1 - m2(g

Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8

Fp = 2.37 N

Fp = 2.37 N

F = 90 degrees

c) When the pin does not exert a force,

Fp = 0

F1 + F2 - Fr = F5

m1 + m2 - mr = m5

m5 = 0.281 + 0.177 - 0.158

m5 = 0.3 kg

Since the net torque on the rod is zero,

t1 + t2 - tr - t5

t5 = t1 + t2 - tr - t5

t5 = t1 + t2 - tr - t5

r5 = r1F1 + r2F2 - ffFr)/F5

r5 = (r1m1 + r2m2 - rrmr)/m5

r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3

r5 = 0.36