A smaller student of mass m= 40 kg hangs from a light rope which passes over a large but massless pulley. The other end of the rope is held by a larger student of mass M= 50 kg who is in contact with the floor. a)With the smaller student hanging at rest, what is the force that the floor exerts on the larger student? For parts (b) and (c), the smaller student climbs up the rope so that he accelerates at 1.9 m/s2.
What is the tension in the rope as the smaller student climbs and accelerates?
While the smaller student climbs and accelerates, what is the force that the floor exerts on the larger student?
d)What minimum acceleration must the smaller student have if he is to cause the larger student to lose contact with the floor?
a) To determine the force that the floor exerts on the larger student when the smaller student is hanging at rest, we can use Newton's second law: F = ma, where F is the force, m is the mass, and a is the acceleration.
Since the smaller student is hanging at rest, there is no acceleration (a = 0). Therefore, the net force acting on the smaller student is zero. The tension in the rope due to the weight of the smaller student is balanced by the force exerted by the floor on the larger student.
We can calculate the force exerted by the floor using the equation: F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).
So, the force that the floor exerts on the larger student is F = M * g = 50 kg * 9.8 m/s² = 490 N.
b) To calculate the tension in the rope as the smaller student climbs and accelerates, we can consider the forces acting on the system. Since the pulley is massless, the tension in the rope is the same on both sides.
The tension in the rope can be calculated using the equation: T = (M + m) * a, where T is the tension, M and m are the masses, and a is the acceleration.
In this case, the tension in the rope is T = (M + m) * a = (50 kg + 40 kg) * 1.9 m/s² = 182 kg m/s² = 182 N.
c) While the smaller student climbs and accelerates, the force that the floor exerts on the larger student can be calculated using Newton's second law. The net force acting on the larger student is the difference between the force due to the tension in the rope and the force due to the larger student's weight.
The force due to the larger student's weight can be calculated using the equation: F = M * g = 50 kg * 9.8 m/s² = 490 N.
So, the force that the floor exerts on the larger student is given by: F = T - M * g = 182 N - 490 N = -308 N. The negative sign indicates that the force is acting in the opposite direction to the force due to the larger student's weight.
d) To determine the minimum acceleration required for the larger student to lose contact with the floor, we need to calculate the force exerted by the floor on the larger student when this happens.
When the larger student loses contact with the floor, the force exerted by the floor becomes zero. This means that the force due to the larger student's weight is balanced by the tension in the rope.
The force due to the larger student's weight is F = M * g = 50 kg * 9.8 m/s² = 490 N.
Therefore, the minimum acceleration required for the larger student to lose contact with the floor can be calculated using the equation: F = (M + m) * a, where F is the force, M and m are the masses, and a is the acceleration.
Rearranging the equation, we can solve for the minimum acceleration: a = F / (M + m) = 490 N / (50 kg + 40 kg) = 490 N / 90 kg = 5.44 m/s².