Question

Figure 5-33 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 98 N on the wall to which it is attached. The tensions in the three shorter cords are T1 = 58.8 N, T2 = 49.0 N, and T3 = 9.8 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D?

Answer 1

is it newtons/gravity?

Answer 2

well?

Answer 3

\[T_{3} = m_{D}g\]\[m_{D} = \frac{T_{3}}{g} = \frac{9.8N}{9.8 m/s} = 1kg\]\[T_{2} = gm_{C}+gm_{D}\]\[m_{C} = \frac{T_{2}-gm_{D}}{g} = \frac{49N-(9.8m/s)(1kg)}{9.8m/s} = 4kg\]\[T_{1} = gm_{B} + gm_{C} + gm_{D}\]\[m_{B} = \frac{T_{1}-gm_{C}-gm_{D}}{g} = \frac{58.8N - (9.8m/s)(4kg) - (9.8m/s)(1kg)}{9.8m/s} = 1kg\]\[98N = gm_{A} + T_{1} +T_{2} + T_{3}\]Everything seemed fine util this....\[m_{A} = \frac{98N - T_{1}-T_{2}-T_{3}}{g} = \frac{98N-58.8N-49N-9.8N}{9.8m/s} = -2kg\]

Ultimately the problem boils down to this: it says the total force pulled on the wall is 98N, however, the 3 given tensions themselves add up to 117.6N, which is already more than it should be.

Answer 4

well the answers are (a) 4.0 kg; (b) 1.0 kg; (c) 4.0 kg; (d) 1.0 kg

Answer 5

Oh wait, I did that last part wrong. It should be like this:\[98N = gm_{A} + T_{1}\]\[m_{A} = \frac{98N - T_{1}}{g} - \frac{98N - 58.8N}{9.8m/s} = 4kg\]

Answer 6

Yeah, that's what I got

Answer 7

awesome thanks!