Need physics help: Dynamics

  • Thread starter Atomos
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  • #1
Atomos
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Note: I am using 9.8 as accleration of gravity; don't ask why
First question is this:

"The drawing shows a circus cloiwn who weighs 890N. The coefficient of static friction between the clown's feet and the ground is 0.53. He pulls down vertically on a rope that passes through 3 pulleys. What is the minimum pulling force required to pull his feet out from beneath him?"

http://img.photobucket.com/albums/v81/uranium235/clown.jpg [Broken]

I get about 308N, am I correct?

second question:

"A mountain climber is in the process of crossing between two cliffs by rope, pauses to rest. She weights 535N. As the drawing shows, she is closer to the left cliffthan to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Determine the tension on the left and the right sides of the rope."

http://img.photobucket.com/albums/v81/uranium235/climber.jpg [Broken]

I speculated that both ropes share the vertical component equally, however that got me nowhere.
edit: oh, and the cliffs are supposed to be of equal height above the climber.

third question:
"In the drawing, the rope and the pulleys are massless, and there is no friction. Find (a) the tension in the rope and (b) the acceleration of the 10kg block."

http://img.photobucket.com/albums/v81/uranium235/pulleys.jpg [Broken]

I speculated that the actual force exerted on the 10kg mass was half the gravitational force on the 3kg mass, and thus the acceration of the system would be the acceleration of the 10kg mass. This reasoning is incorrect, why?
 
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Answers and Replies

  • #2
Doc Al
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Your answer to the first problem seems correct, but you didn't show your work.

As far as the other problems, why all the speculation? Label the forces and figure it out. For the mountain climber problem, there is equilibrium, so the net force on the climber must be zero. For the pulley and masses, there is acceleration. (Hint: the rope tension is the same throughout.)
 
  • #3
Atomos
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work to first problem:
[tex]F_f = uF_n[/tex]
[tex]F_n = |F_g| - |F_a|[/tex]
[tex]|F_a| > u|F_g| - u|F_a|[/tex]
[tex] F_a( 1 + u) > 890u[/tex]
[tex] F_a > ~308.3[/tex]
 
  • #4
Atomos
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OK, so for the second one, the horizontal components of the tension on each rope should be equal and opposite right? And the vertical components should add up to 535N but not necessarily be 1/2 of 535N?
 
  • #5
Atomos
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[y]es/[n]o? Please, I may be tested on this tomorrow.
 
  • #6
Doc Al
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Atomos said:
work to first problem:
[tex]F_f = uF_n[/tex]
[tex]F_n = |F_g| - |F_a|[/tex]
[tex]|F_a| > u|F_g| - u|F_a|[/tex]
[tex] F_a( 1 + u) > 890u[/tex]
[tex] F_a > ~308.3[/tex]
Looks OK to me.
 
  • #7
Doc Al
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Atomos said:
OK, so for the second one, the horizontal components of the tension on each rope should be equal and opposite right? And the vertical components should add up to 535N but not necessarily be 1/2 of 535N?
Exactly right.
 

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