Question

A uniform rod of mass mr = 173 g and length L = 1... A uniform rod of mass mr = 173 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 = 246 g and m2 = 127 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 =? m Fp =? N ?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 =? kg Fp =? N ?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 locationr5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 =? kg r5 =? m

Answer 1

a)

T₁ = tension in string 1 at r₁ = m₁g = 0.246 x 9.8 = 2.41 N

T₂ = tension in string 2 at r₂ = m₂g = 0.127 x 9.8 = 1.24 N

using equiilbrium of torque

T₁ r₁ + T₂ r₂ = m_r g (L/2) + m₃g r₃

2.41 (0.10) + 1.24 (0.90) = (0.173) (9.8) (0.50) + (0.2)(9.8)r₃

r₃ = 0.26 m = 26 cm

F_x = horizontal force by pin

using equilibrium of force in horizontal direction

F_x = 0

using equilibrium of force in vertical direction

F_y + T₁ + T₂ = m_r g + m₃g

F_y + 2.41 + 1.24 = (0.173) (9.8) + (0.2) (9.8)

F_y = 0.0054 N