Question

The drawing shows Robin Hood (mass = 86 kg) about to escape from a dangerous situation. As you can see, currently the chandelier (mass = 226 kg) is held 2.5 m off the floor by a rope looped over the beams and tied to the floor. When Robin cuts the rope free from the floor with one hand, he will hold on to the rope with the other. He will then be pulled safely up to the balcony 2.1m above as the chandelier falls. (Ignore the friction between the rope and the beams over which is slides and take a co-ordinate system where up is positive y.)
1. What is the magnitude of acceleration at which Robin Hood pulled upwards after the rope has been cut? |aRH| =

Answer 1

Answer:

The answers to the question are

a). a = 4.402 m/s^2

b). T = 1222.23 N

Explanation:

Applying Newton's second law we have

For Robin Hood, T - mg = ma and for the chandelier, T - Mg = -Ma

Please note that the upwards motion is assumed positive therefore, upward acceleration, a for Robin Hood is +ve and that of the chandelier and gravity acceleration are -ve

Solving Robin Hood's equation for T, we have

T = mg + ma substituting the value for T in the chandeliers equation we have

mg+ma-Mg=-Ma or a = (M-m)×g/(M+m)

Therefore, Robin Hood's acceleration (M-m)×g/(M+m) = (226 kg-86 kg) × (9.81 m/s^2)/((226 kg)+(86 kg)) = 4.40 m/s^2

b). Substituting the value to solve for T = mg + ma = 86×(9.81+4.4) = 1222.23 N