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

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). PLEASE HELP ME FIND FP. The answer isn't 1.9 or 3.2, .8, .6, .58, 4.8 or .4.

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). I JUST NEED FP FOR THIS AS WELL. The answer for this Fp isn't 3.2, 1.96, 4.8 or .05.

Explanation / Answer

Part a )

given

mr = 118 g

length L = 100.0 cm

r1 = 10.0 cm and

r2 = 90.0 cm

m1 = 231 g and m2 = 147 g , m3 = 200 g

r3 = ( r1 m1 + r2 m2 - (L/2)mr ) / m3

r3 = 0.1 X 0.231 + 0.9 X 0.147 - 0.5 X 0.118 / 0.2

r3 = 0.482 m

or

r3 = 48.2 cm

Fp = (-m1g-m2g+mrg)

Fp = - 0.231 X 9.8 - 0.147 X 9.8 + 0.118 X 9.8

Fp = - 2.548 N

Part b)

mr = 118 g

length L = 100.0 cm

r1 = 10.0 cm and

r2 = 90.0 cm

m1 = 231 g and m2 = 147 g ,

r4 = 20.0 cm = 0.2 m

m4 = (r1m1+r2m2-(L/2)mr )/ r4

m4 = (0.1X0.231+0.9X0.147-0.5X0.118)/0.2

m4 = 0.482 Kg

m4 = 482 g

Fp = (-m1g-m2g+mrg+m4g)

Fp = (-0.231 X 9.8 - 0.147 X 9.8 + 0.118 X 9.8 + 0.482 X 9.8)

Fp = 2.1756 g

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