A rope of negligible mass passes over a puley of a negligible mass attached to the ceiling. One end of the rope is held by stdent A of mass 70 kg who is at rest on the floor. The opposite end of the rope is held by student B of mass 60 kg who is suspended at rest above the floor
calculate the magnitude of the force exerted on the floor by student A
I said 700 N and aparently this is wrong and I don\'t see why
The suspended student B's weight is balanced by a rope force Mb*g = 588 N. This becomes an upward force on Student A. A is held in equilibrium by ground force Fg (up), rope force (up) and his weight Mb*g (down).
Fg + 588 = Mb*g = 686 N
Fg = 686 - 588 = 98 N
If you are using 10 m/s^2 for g, you'd get 100 N for the answer. I used 9.8.
To calculate the magnitude of the force exerted on the floor by student A, you need to analyze the forces acting on the system.
In this scenario, the forces acting on the system are the weight of student A (mg) and the tension force in the rope (T). Since the system is in equilibrium (at rest), the net force acting on the system must be zero.
Considering the forces acting on student A, we have:
1. Weight of student A (mg): The weight of student A can be calculated by multiplying their mass (70 kg) by the gravitational acceleration (9.8 m/s²). Therefore, the weight of student A is given by W = mg = 70 kg * 9.8 m/s² = 686 N.
Since the system is in equilibrium, the tension in the rope (T) must be equal in magnitude to the weight of student A, but acting in the opposite direction. Therefore, the magnitude of the force exerted on the floor by student A is 686 N.
Apologies for the mistake in my initial response where I stated 700 N as the magnitude of the force exerted on the floor. The correct value is 686 N.