Force and Motion (with tension)

[th77]

Holding on to a towrope moving parallel to a frictionless ski slope, a 50 kg skier is puilled up the slope, which is at an angle of 8 degrees with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.0 m/s and (b) v = 2.0 m/s as v increases at a rate of 0.10 m/s^2?
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I drew the free body diagram. I have the normal force Fn along the y-axis. I have a force T (tension) along the x-axis. Then I have Fg making an angle that is 8 degrees to the negative y-axis.
My first question: Did I draw the FBD correctly?
Next question: How do I proceed from here?

 

[GregA]

This may not be of much use as I have not done too much work yet with mechanics but shouldn't the tension force be parallel to the ski slope and acting upwards? as opposed to aligned with the x-axis, the persons weight has two components, and in the original question -part (a)- as velocity is constant and friction is ignored, one of the vectors that that represents the person's tendency to move down the slope should be equal and opposite to the tension in the string.

(the normal force should be of help for finding the frictional force but here it is nil)

 

[th77]

This may not be of much use as I have not done too much work yet with mechanics but shouldn't the tension force be parallel to the ski slope and acting upwards?

I'm not sure. Thats part of my question.

 

[GregA]

The person being pulled up the ski slope resides on the same plane as the object that is towing him (or her). Imagine where the rope would be if connected to these two bodies, what is causing tension in the rope? and which of these is relevant to your question?

 

[Doc Al]

I drew the free body diagram. I have the normal force Fn along the y-axis. I have a force T (tension) along the x-axis. Then I have Fg making an angle that is 8 degrees to the negative y-axis.
My first question: Did I draw the FBD correctly?
Next question: How do I proceed from here?

Assuming your x and y axes are defined as parallel and perpendicular to the slope, it looks like you are on the right track. Next consider the forces parallel to the slope (along the x direction), as that's the only direction that the skier can move. (What's the component of the weight along the x-direction?) Apply Newton's 2nd law to both cases to solve for the tension force.

 

[th77]

Assuming your x and y axes are defined as parallel and perpendicular to the slope, it looks like you are on the right track. Next consider the forces parallel to the slope (along the x direction), as that's the only direction that the skier can move. (What's the component of the weight along the x-direction?) Apply Newton's 2nd law to both cases to solve for the tension force.

The weight along the x-direction is mgsin 8 . Does it matter what the y component of weight is?

Using Newtons 2nd law would it be T - mg sin 8 = ma ?

 

[Doc Al]

The weight along the x-direction is mgsin 8 .

Right.

Does it matter what the y component of weight is?

Not to me. (All we care about are forces in the x-direction. If there was friction, then the normal force would matter.)

Using Newtons 2nd law would it be T - mg sin 8 = ma ?

Exactly.

 

[th77]

Thanks very much!