Raphael K. answered • 09/17/21
BIG BRAIN : Chemical Engineer teaching - Chemistry, Math, and Physics.
A meter ruler is pivoted
A meter ruler is pivoted about the '50cm' mark (which is the position of its centre of mass). There is a 100g mass sitting on the 20cm mark, and a 50g mass sitting on the 85cm mark. Force F, acts at the 35cm mark. Calculate the direction and magnitude of force F needed to balance the rule. Assume the upwards direction is positive, and take g=9.8Nkg-1
+τ,100g +τ,F - τ,50g
0—10—20—30—40—50—60—70—80—90—100 (cm)
Δ
Torque = ⊥Force * distance.
Torque is a vector that operates in the 3rd plane using the right hand rule. Look it up or ask if you're not sure about the right hand rule.
Using the distance from the fulcrum, or center of gravity of the ruler, find the sum of the torques from each mass and set them equal to 0.
Using Right-Hand-Rule Orientation: CCW = +τ
0 = +(.30m)*(-9.81m/s2)(.100kg) + (0.15m)F - (0.35m)(-9.81m/s2)(0.050kg)
0 = -0.2943 kg*m2/s2 + 0.15m* F + 0.172 kg*m2/s2
0.1223 kg*m2/s2 / 0.15m = F
F = 0.815 kg*m/s2
F = 0.815 N
