1. Three balls are attached to a light rigid rod, as shown in the figure. A fulc

Question

1. Three balls are attached to a light rigid rod, as shown in the figure. A fulcrum is located at xf = 19.5 cm measured from the left end of the rod. Ball A has mass mA = 60 g and is located at xA = 9 cm. Ball B has mass mB = 9 g located at xB = 23 cm. Ball C has mass mC = 22 g located at xC = 37 cm. Calculate the x-coordinate of the center of mass XCM. Assume that the mass of the rod is small enough to be ignored in your calculation. Your answers should be accurate to within 0.1 cm

XCM = ? cm

2. In Part 1 above, is the structure in balance? If not, how far must the fulcrum be moved so that it becomes in balance? (Answer "0.0" if it does not need to be moved. Also, signs matter: + if to the right,

Explanation / Answer

Taking the end of the rod nearer to A as the origin to calculate the center of mass
1) Xcm = 1/M ( xi * mi) = (xa ma + xb mb + xc mc )/(ma+mb+mc) = (9*60+23*9 + 22*37)/(60+9+22)=17.15 cm

Xcm = 17.2 cm (approximated nearest to .1cm) from the end closed to A

2) lets assume the fulcrum is moved by a distance x away from A

NetTorque must be equal for the system to be in equilibrium

ma (xf+x - xa) = mb(xb-(xf+x) ) +mc(xc-(xf+x))

60(19.5 +x - 9) = 9(23 -19.5-x) +22(37-19.5 -x)

60x+ 630 = 31.5 - 9x + 385 -22x

91x = 213.5

x= -2.346   

xfnew= xf +x = 19.5 - 2.346 = 17.154 = 17. 2 cm

The the fulcrum needs to be moved 2.3 cm towards A for the system to stay in equilibrium.

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