Asked by Sam
Figure 5-53 shows a man sitting in a bosun's chair that dangles from a massless, frictionless pulley and back down to the mans's hand. The combined mass of man and chair is 94 kg.
Fig. 5-53
With what force magnitude must the man pull on the rope if he is to rise
(a) with a constant velocity? 1 N
(b) with an upward acceleration of 1.27 m/s2? 2 N
If the rope extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the man to rise
(c) with a constant velocity? 3 N
(d) with an upward acceleration of 1.27 m/s2? 4 N
Fig. 5-53
With what force magnitude must the man pull on the rope if he is to rise
(a) with a constant velocity? 1 N
(b) with an upward acceleration of 1.27 m/s2? 2 N
If the rope extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the man to rise
(c) with a constant velocity? 3 N
(d) with an upward acceleration of 1.27 m/s2? 4 N
Answers
Answered by
Slevin Kelevera
Ok since I had to search for bits and pieces of how to solve this problem online for 2 hrs I'm gonna put the answer here.
My combined mass was 102.0 kg
for b n c my accel was 1.23 m/s^2
First off, there are 3 forces working in this free body diagram, mg*g (downward/neg) and T on both sides of the pulley, T is the same for both sides.
Part a. F=m*a (accel=0)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
0=-999.6+2T - solve for T
Part b. F=m*a (accel=1.23)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
102.0*1.23=-999.6+2T - solve for T
** T changes from T to T/2 because of the guy standing on the ground. ** Find a pic of a pulley with one side of the rope attached to the ground or ceiling and you will understand.
Part c. F=m*a (accel=0)
Fnet=-(m*g)+2(T/2)
Since it's constant Velocity, Accel=0
0=-999.6+2(T/2) - solve for T
Part d. F=m*a (accel=1.23)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
102.0*1.23=-999.6+2(T/2) - solve for T
** I had a part e, What is the magnitude of the force on the ceiling from the pulley system in part a, part b, part c, and part d?**
Part e.
Just multiply values by 2 because there are 2 tensions on the pulley. Simple as that. Hope other people find this and don't waste the time I spent looking for everything. I do have to say I fully understand this now though.
My combined mass was 102.0 kg
for b n c my accel was 1.23 m/s^2
First off, there are 3 forces working in this free body diagram, mg*g (downward/neg) and T on both sides of the pulley, T is the same for both sides.
Part a. F=m*a (accel=0)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
0=-999.6+2T - solve for T
Part b. F=m*a (accel=1.23)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
102.0*1.23=-999.6+2T - solve for T
** T changes from T to T/2 because of the guy standing on the ground. ** Find a pic of a pulley with one side of the rope attached to the ground or ceiling and you will understand.
Part c. F=m*a (accel=0)
Fnet=-(m*g)+2(T/2)
Since it's constant Velocity, Accel=0
0=-999.6+2(T/2) - solve for T
Part d. F=m*a (accel=1.23)
Fnet=-(m*g)+2T
Since it's constant Velocity, Accel=0
102.0*1.23=-999.6+2(T/2) - solve for T
** I had a part e, What is the magnitude of the force on the ceiling from the pulley system in part a, part b, part c, and part d?**
Part e.
Just multiply values by 2 because there are 2 tensions on the pulley. Simple as that. Hope other people find this and don't waste the time I spent looking for everything. I do have to say I fully understand this now though.
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