Question: I need help with this asap. I'm on my last attempt and need to get correct. Can you please show all the answers neatly so

I need help with this asap. I'm on my last attempt and need to get correct. Can you please show all the answers neatly so I can see them correctly.Thank you. 12a

I need help with this asap. I'm on my last attempt and

As a torque activity, your Physics TA sets up the arrangement shown below. A uniform rod of mass m, = 113 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, = 10.0 cm and r2 - 90.0 cm mark, passed over pulleys, and masses of m, = 281 g and my = 177 9 are attached. Your TA asks you to determine the following (a) The position r; on the rod where you would suspend a mass ma = 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 ,, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). (b) Let's now remove the mass my and determine the new mass my you would need to suspend from the rod at the position , - 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 ,, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). ma = J kg F. = IN (c) Let's now remove the mass m, and determine the mass m, 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 ( from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. ms = kg

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