In gym class a student hangs on a rope connected to a force probe. If he is at r
ID: 1536775 • Letter: I
Question
In gym class a student hangs on a rope connected to a force probe. If he is at rest, the force probe measures a value of 525 N. If the rope is 10.0 m long and he slides down from the top with a uniform acceleration In a time of 2.50 s, what will the force probe measure during his downward motion? A 25.0 kg mass is hanging from two pulleys as shown at right. If the system is stationary, what are the forces: T_1, T_2, T_3, T_4, T_5, and the force F which must be pulled to maintain this state? (assume pulleys are frictionless and masses, and rope is massless too) 3 blocks (A, B, and C) have masses 5.00 kg, 10.0 kg, and 15.0 kg, respectively. What Force, F, is needed to maintain a constant acceleration of 2.50 m/s2 for this system? (assume frictionless surface) ALSO, what is the force between each block, F_ab and F_bc?Explanation / Answer
Given; mass m = 25.0 kg
Looking only at m and its rope and summing vertical forces to zero.
T1 = m*g
= 25.0 kg * 9.8 m/s^2
T1 = 245 N
Looking only at the lower pulley and summing vertical forces to zero.
T2 = T3 = m*g/2
T2 = T3 = 25.0*9.8/2
T2 = T3 = 122.5 N
as the pulleys are frictionless
T2 = T3 = T4 = m*g/2
= 122.5 N
F = T4 = m*g/2
F = 122.5 N
Looking only at the upper pulley and summing vertical forces to zero.
T5 = 3*m*g/2
= 3*25.0*9.8/2
= 367.5 N