a) The position r 3 on the rod where you would suspend a mass m 3 = 200 g in ord
ID: 2040240 • Letter: A
Question
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 =
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).
Fp =?
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 =
r3 =
Fp =
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 =?
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 =
Explanation / Answer
tension in left cable T1 = m1*g
tension in right cable T2 = m2*g
m1 = 281 g = 0.281 kg
m2 = 157 g = 0.157 kg
m3 = 200 g = 0.2 kg
mr = 148 g = 0.148 kg
r1 = 10 cm = 0.1 m
r2 = 90 cm = 0.9 m
L = 100 cm = 1 m
r3 = ?
In equilibrium net torque = 0 about the pivot
T1*r1 + T2*r2 - mr*g*L/2 - m3*g*r3 = 0
m1*g*r1 + m2*g*r2 - mr*g*L/2 - m3*g*r3 = 0
m1*r1 + m2*r2 - mr*L/2 - m3*r3 = 0
0.281*0.1 + 0.157*0.9 - 0.148*1/2 - 0.2*r3 = 0
r3 = 0.477 m = 47.7 cm <<<-----------ANSWER
In equilibrium
along vertical
Fnet = 0
Fp + T1 + T2 - mr*g - m3*g = 0
Fp + m1*g + m2*g - mr*g - m3*g = 0
Fp = (mr + m3 - m1 - m2)*g
Fp = (0.148 + 0.2 - 0.281 - 0.157)*9.8
Fp = -0.882 N
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(b)
In equilibrium net torque = 0 about the pivot
T1*r1 + T2*r2 - mr*g*L/2 - m4*g*r4 = 0
m1*g*r1 + m2*g*r2 - mr*g*L/2 - m4*g*r4 = 0
m1*r1 + m2*r2 - mr*L/2 - m4*r4 = 0
0.281*0.1 + 0.157*0.9 - 0.148*1/2 - m4*0.2 = 0
r3 = 0.477 m = 47.7 cm <<<-----------ANSWER
m4 = 0.477 kg = 477 g
In equilibrium
along vertical
Fnet = 0
Fp + T1 + T2 - mr*g - m4*g = 0
Fp + m1*g + m2*g - mr*g - m4*g = 0
Fp = (mr + m4 - m1 - m2)*g
Fp = (0.148 + 0.477 - 0.281 - 0.157)*9.8
Fp = 1.8326 N
======================================
( c )
In equilibrium
along vertical
Fnet = 0
Fp + T1 + T2 - mr*g - m3*g = 0
Fp + m1*g + m2*g - mr*g - m5*g = 0
for Fp = 0
m5 = m1 + m2 - mr
m5 = 0.281 + 0.157 - 0.148
m5 = 0.29 kg
m5 = 290 g
In equilibrium net torque = 0 about the pivot
T1*r1 + T2*r2 - mr*g*L/2 - m5*g*r5 = 0
m1*g*r1 + m2*g*r2 - mr*g*L/2 - m5*g*r5 = 0
m1*r1 + m2*r2 - mr*L/2 - m5*r5 = 0
0.281*0.1 + 0.157*0.9 - 0.148*1/2 - 0.29*r5 = 0
r3 = 0.329 m = 32.9 cm = 33 cm <<<-----------ANSWER
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