Physics question : mobile, static equilibrium?

heres this picture of the mobile


http://i28.tinypic.com/24qogzq.jpg





a designmer wishes to construct a mobile using strings and light rods (neg. mass)


if the mass on the bottom left is 125 g


what are the masses of the other figures so that rod remins in horizontal?

Physics question : mobile, static equilibrium?
Start with the bottom rod. ΣM = 0 → 30*125 = 20*a → a = 187.5. The total weight of the lowest rod is therefore 125+187.5 = 312.5





Use this weight for the next higher rod: 312.5*10 = 30*b. Solve for b and add to 312.5 to get the input for the highest rod.





By now you see the pattern. Step by step.....
Reply:Just start balancing your torques for each section of the mobile starting at the bottom. Having a net torque of 0 produces a state of rotational equilibrium meaning its gonna stay horizontal.


The equation for calculating torque is:


t = r F sin(Angle)


where:


t, is the torque


r, is the distance from pivot point the force is being applied


F, is the magnitude of the force being applied


Angle, the angle at which the force is being applied and since in this case case always going to be 90 degrees and sin(90) = 1 the equation can be simplified to just:


t = r F





Start at the bottom


rotational equilibrium like i said before is achieved when the net t is zero. so:


t net = 0


but the tnet is being provided by 2 torques the 125g weight on the left (t1) and the unknown one on the right (t2). so:


t1 - t2 = 0


F1 r1 - F2 r2 = 0


since the only force in this case is being provided by the weight of the weights then your F = mg


m1g r1 - m2g r2 = 0


m1g r1 = m2g r2


m1 r1 = m2 r2


m2 = [ m1 r1 ]/[ r2 ]


now looking at this equation you know everything except for the value of m2 plug in and solve for it.


m2 = [ (.125kg)(.3m) ]/[ .2m ]


m2 = .1875 kg = 187.5g


That would be the mass of the weight label A.


The rest of the problem is just a repetition of this except that at the next layer the force acting at the 10cm side of the pivot is going to be the combine weight of the 2 bottom weights ( 125g + 187.5g = 312.5g ) and at the level above that its going to be the combines weight of the bottom 3 weights.


After doing the math you should get


weightC = 250g


weightB = 104.2g


weightA = 187.5g