Masses of 100 and 500 grams are placed at 0 cm and 100 cm points of a meter stic
ID: 1495924 • Letter: M
Question
Masses of 100 and 500 grams are placed at 0 cm and 100 cm points of a meter stick respectively. Where must a single vertical force be placed to achieve a balance What is the value of F in Newtons The meter stick is uniform and has a mass of 80 grams. Position of F__________cm, Magnitude of F__________Newton A meter stick whose mass is 0.200 kg is supported at the zero cm mark by a knife edge and a force F at the 100 cm point. A mass of 700 grams is attached to the stick at the 40 cm mark. Find the magnitude of N and F in Newtons. Magnitude of N__________Newton, Magnitude of F________NewtonExplanation / Answer
1) let m1 = 100 g = 0.1 kg
m2 = 500 g = 0.5 kg
let x is the distrance from left end where F should be applied.
Apply net torque about the pint where F is applied.
Tnet = 0
m1*g*x - m2*g*(100 - x) = 0
m1*g*x - m2*g*100 + m2*g*x = 0
x = m2*g*100/(m1*g + m2*g)
= m2*100/(m1 + m2)
= 0.5*100/(0.1 + 0.5)
= 83.3 cm
F = (m1+m2)*g
= (0.1 + 0.5)*9.8
= 5.88 N
2) Apply net torque about right = 0
N*1 - 0.7*9.8*0.6 - 0.2*9.8*0.5 = 0
N = 0.7*9.8*0.6 + 0.2*9.8*0.5
= 5.096 N
Apply, Fnety = 0
N - 0.7*9.8 - 0.2*9.8 + F = 0
F = 0.7*9.8 + 0.2*9.8 - 5.096
= 3.724 N