A uniform metre rule of mass 90g is pivoted at the 40cm mark. If the rule is in equilibrium with an unknown mass m placed at the 10cm mark and a 72g mass at the 70cm mark,determine m.
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Assume all of the mass of the meter ruler is centered at center of gravity at 50 cm.
Sum of the torques must be zero
Pivot is at 40 cm
Torque due to mass of ruler = 10 * 90 = 900 g cm
Torque due to 72 g = 30 * 72 = 2160 g cm
Torque due to M = M * 30 and opposite direction
30 M = 2160 + 900 = 3060
M = 102 g
Sum of the torques must be zero
Pivot is at 40 cm
Torque due to mass of ruler = 10 * 90 = 900 g cm
Torque due to 72 g = 30 * 72 = 2160 g cm
Torque due to M = M * 30 and opposite direction
30 M = 2160 + 900 = 3060
M = 102 g
it's understood now
How did you get 30 in the solving for torque
The distance between the point of pivot and the mass is 30cm
Don't forget perpendicular distance
Don't forget perpendicular distance
Ali