3 physics question help

  • #1
If anyone can help this physics novice out, I have been working on these problems for about a week. I just started looking at my old engineering Physics books and physics books, trying to brush up.... so if you can help, here are the 3 questions:

1. A sled of mass 10 kg is held in place on a frictionless 25 degree slope by a rope attached to a stake at the top. The rope is parallel to the slope. What is the tension in the rope.

2. An elevator weighing 25,000 N is supported by a steel cable. What is the tension in the cable when the elevator is being accelerated upward at a rate of 2 m/s^2. (g = 9.8 m/s^2)

3. A Ferris wheel has radius 5.0 m and makes one revolution in 8 seconds. A person weighing 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion?

Answers and Replies

  • #2
Have you tried drawing free body diagrams? They'll help a lot in problems like these.
  • #3
(I have to put the diagram into words because I haven't figured out how to put an image in this message)
Would the free-body diagram for the elevator have tension going upwards, (N) going downwards, and (mg) going downards. Another piece of info I do not get is how would the acceleration which is going upwards fit into the force diagram. If I am missing anything for this diagram please correct me and please correct me if my equation is wrong.
Would the equation look like this T - N -mg = ma ?

For the sled, the free body diagram would be for the vertical N upward, -F[size=.6]g[/size]downward, and for the horizontal tension to the right. My question is does the stake have a free body diagram?
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  • #4
I'll try and assist you in constructing the free body diagram.

For the elevator question, here's a hint: according to Newton's 3rd Law, the force on the elevator from the cable is equal to the tension in the cable.

The force on the elevator due to the cable is pulling it upwards, while the elevator's own weight is pulling it downwards. These two forces will add up to give a net force which causes the upward acceleration.

For the sled, three forces are acting on it. The tension from the rope (parallel to slope), the weight (vertically downwards) and the normal reaction force (which is always perpendicular to the surface). There is no net force.

Hopefully this will help you construct the necessary diagrams. You can actually attach images to this, and it's easy for me to draw images using Paint and show you, but hanging around this forum tells me that I should not give answers directly.
  • #5
Would the diagram for the elevator and sled look like this:

Also the tension (t) should be going downwards, Force (F) should be going upwards

( image should be on page two) I accidently deleted off this page ,and now it will not allow me to post on this page anymore.)
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  • #6
sorry the image is on this page
Also here is what I have for the sled.

T = F1g(sin25)

F1g = (10 kg)(9.8m/s^2)
F1g = 98 N
plug into the equation T = F1g(sin25)

T = 98 N(sin25)
T = 41.41 N

Wouldn't Fxg= Fgsin 25 and Fyg = Fgcos 25

Also could you please help me on the setup with the sled. I am kinda confused on what I would do next.

Now back to the elevator problem, (I think my force diagram is incorrect because doesn't force and tension have to be in opposite directions (the force(acceleration) is upwards and the tension is upwards would that be correct?))

First equation T-F1g = m (a) ( I think my setup is correct this time around. If not please tell me what I am doing incorrectly?
T-25,000 N = 2,551kg(2 m/s)
T -25,000 N = 5,102 N
T = 25,000 N +5,102N
T =30,102 N

F1g=25,000 N ( would this be correct)
Mass would equal 25,000 N / 9.8 m/s^2 which equal 2551 kg
My next question would be is the mass correct and did I set it up correctly?


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