Suppose that a 100cm meter stick is balanced at its center 50cm. A 0.24 kg mass

Question

Suppose that a 100cm meter stick is balanced at its center 50cm. A 0.24 kg mass is then positioned at the 6 cm mark. At what cm mark must a 0.35 kg mass be placed to balance the 0.24 kg mass?

Given the situation in the figure. The mass m1 is 0.54 kg and it is located at x1 = 25 cm. The pivot point is represented by the solid triangle located at x = 60 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.24 kg and it is located at x2 = 85 cm. Calculate the net torque (in N·m with the proper sign) due to these three weights. Use g = 9.79 m/s2.

Cnt post the picture but its looks just as explained above

Explanation / Answer

I assume the two questions are completely separate. For the first one you need to use the formula for torque T = F*r (this is a simplified formula since in this case all the forces are perpendicular). Find the torque by induced by the first mass. r is the distance from the center (so 50cm - 6 cm). The force F is the force of gravity on mass 1 F = mg. Now it says the rod is balanced so that means the torques must cancel out (be equal). So you have the torque caused by mass 1, use it in the same equation to solve for the distance (r2) mass 2 needs to be from the center to create the same torque. The Force term F is now the force of gravity on mass 2. This distance needs to be added to the 50cm to find the location on the meter stick. For the second question use the same formula for torque to calculate the values for m1, m2, and the meter stick. (Remember to use the distance from the pivot point not not the center). Give m2 (on the right of the pivot point) a positive sign and the other two (on the left) negatives. This Is because the forces want to cause the meter stick to rotate in opposite directions. Add everything together and there is your overall torque.

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