Asked by Jack
Consider a heavy rope with mass per unit length 2.10 kg/m; a length L=18.00 m of the rope is hanging vertically from a support, as shown. If the support and rope are placed in an elevator accerlerating downward with a=0.800 m/s^2, what is the tension in the rope at point P? (at point P, mass is 12.6)
This is what I did:
F=ma
F=(12.6)(0.8)
F=10.08
Sum of F = T-mg = 10.08 N
Sum of F = T-(12.6)(9.81) = 10.08 N
Therefore, T is equal to 134 N.
It was the wrong answer, could someone help me?
This is what I did:
F=ma
F=(12.6)(0.8)
F=10.08
Sum of F = T-mg = 10.08 N
Sum of F = T-(12.6)(9.81) = 10.08 N
Therefore, T is equal to 134 N.
It was the wrong answer, could someone help me?
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