The figure shows an arrangement in which four disks are suspended by cords. The

Question

The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 95.1 N on the wall to which it is attached. The tensions in the shorter cords are T_1 = 67.0 N, T_2 = 32.1 N, and T_3 = 8.4 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D? Can't figure out how to load pic. heres the arrangement of the disk though A > T_1 ( is the tension betwen disk A and B) B > T_2 (tension between disk B and C) C > T_3 (tenstion between disk C and D) D

Explanation / Answer

Start with disk D and Force T3
F=ma where a=g=9.8m/s/s
8.4=mD(9.8)
mD= .86kg

Same thing for force T2 but now we have two masses adding to the force, one of those is known
f=ma
T2=(mC+mD)g
32.1=(mC+1)(9.8)
mC = (32.1/9.8) - .86
mC=2.42 kg

same process for T3 only have 3 masses 2 of them known
T3=(mB+mC+mD)g=67

mB=3.56kg

and as a frictionless pulley does not change the magnitude of the force in the cable, only the direction then T4 will be 95.1 N

T4=(mA+mB+mC+mD)g

mA=2.86kg

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