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

Question

A uniform rod of mass mr = 118 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 = 231 g and m2 = 147 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). Find R3, Fp, and thetaF. (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). Find m4, and Fp. Please help! I've been stuck on this for ages.

Explanation / Answer

a) 0.200*r3*g + 0.50*0.118*g = 0.231*0.1*g + 0.147*0.9*g or
r3 = 5*[0.0231+0.1323 -0.059] = 0.482m or 48.2cm
Force = 0.200*9.81 = 1.962 N at an angle of 270 degree from horizontal

b) m4*0.20*g + 0.50*0.138*g = 0.231*0.1*g + 0.147*0.9*g or
m4 = 5*[0.0231+0.1323 -0.059] = 0.482 kg or 482 g

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