you have a meter stick from 0 to 100 with m1 at 20 m2 at 70 and at 50 is a point

Question

you have a meter stick from 0 to 100 with m1 at 20 m2 at 70 and at 50 is a point. these are forces applied by hanging two masses m1 and m2. so with this in mind if mass m1=0.100kg acts at 20.0 cm, what is the value of mass m2 that must be placed at the psition 70.0 cm shown to put the system in equilibrium? Write the equation for sum of torque ccw = sum of torque cw with the mass m2 as unknown and solve for m2. Assume that the meter stick is uniform and symmetric. show work.

Explanation / Answer

Balancing the torque about the point at 50 cm (mid point of meter stick) => 0.1*(50-20) = m2 * (70-50),,,, on solving we get m2= 0.15 kg, Also note that if there is a mass at the midpoint , then also it doesn't have any effect on the torque and the weight of stick is evenly distributed hence that also cancels out in torque balance. Hence answer m2- 0.15 Kg

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