A uniform rod of mass m_r = 123 g and length L = 100.0 cm is attached to the wal

Q: A uniform rod of mass m_r = 123 g and length L = 100.0 cm is attached to the wal

a) solve the r3 value as follows: 0.200*r3*g + 0.50*0.123*g = 0.266*0.1*g + 0.147*0.9*g 0.200*r3 + 0.50*0.123 = 0.266*0.1 + 0.147*0.9 Then, r3 = 0.487 the Force is, F = 0.200*9.81 = 1.962 N at an angle of 270o the value of m4 is calculated as follows m4*0.20*g + 0.50*0.123*g = 0.266*0.1*g + 0.147*0.9*g m4*0.20 + 0.50*0.123 = 0.266*0.1 + 0.147*0.9 the value of m4 is, m4 = 0.487 kg The force is, F = 0.487*9.81 = 4.77747 N = 4.8 N

ID: 1471749 • Letter: A

Question

A uniform rod of mass m_r = 123 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 r_1 = 10.0 cm and r_2 = 90.0 cm mark, passed over pulleys, and masses of m_1 = 266 g and m_2 = 147 g are attached. Your TA asks you to determine the following. The position r_3 on the rod where you would suspend a mass m_3 = 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 Theta_F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). Let's now remove the mass m_3 and determine the new mass m_4 you would need to suspend from the rod at the position r_4 = 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 Theta_F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). Let's now remove the mass m_4 and determine the mass m_5 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 r_5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position.

Explanation / Answer

solve the r3 value as follows:

0.200*r3*g + 0.50*0.123*g = 0.266*0.1*g + 0.147*0.9*g

0.200*r3 + 0.50*0.123 = 0.266*0.1 + 0.147*0.9

Then,
r3 = 0.487
the Force is,

F = 0.200*9.81 = 1.962 N at an angle of 270o

the value of m4 is calculated as follows

m4*0.20*g + 0.50*0.123*g = 0.266*0.1*g + 0.147*0.9*g

m4*0.20 + 0.50*0.123 = 0.266*0.1 + 0.147*0.9

the value of m4 is,

m4 = 0.487 kg

The force is,

F = 0.487*9.81 = 4.77747 N = 4.8 N

Navigate

A uniform rod of mass m r = 118 g and length L = 100.0 cm is attached to the wal

A uniform rod of mass mr = 118 g and length L = 100.0 cm is attached to the wall

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

A uniform rod of mass m_r = 123 g and length L = 100.0 cm is attached to the wal

Question

Explanation / Answer

Related Questions

Navigate