Mariana A.

asked • 12/21/17

what is the magnitude of the pull required to make the crate move up the ramp at a constant velocity?

A 120 N crate is being pulled up a perfectly smooth ramp that slopes upward at 32 degrees by a pull that is directed at 72 degrees above the surface of the ramp. what is the magnitude of the pull required to make the crate move up the ramp at a constant velocity?

Tony H.

tutor
Arturo,
 
I went through the problem again, solving for the normal force, and you are correct - the problem doesn't make sense.  I think I figured out why:
 
With that large of an angle relative to the ramp (72 degrees), we balanced out the forces acting parallel to the ramp - Those forces have magnitudes of 63.6N - They obviously balance each other out since the velocity is supposed to be constant.  However, with that large of an angle (72 degrees), we cannot say that the forces acting perpendicular to the ramp will balance out.  There will be a net force (and hence net acceleration) in the direction perpendicular to the ramp - The crate will be lifted off the ramp up into the air.  The component of the pulling force that is perpendicular to the ramp is 195.7N (as you determined as well), which is more than the weight of the crate (120N) - Hence, the negative "normal" force.
 
So you were correct in the first place - The problem does not work as written... It needs more realistic numbers.
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12/22/17

Arturo O.

Tony,
 
Thank you for that explanation.  Perhaps there are typos in the angles of the problem statement.  In an earlier comment at the bottom of this page, I requested that the student double check the numbers.
 
On another matter, thank you for your service to our country in the Nuclear Navy and the US Strategic Command!  
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12/22/17

Tony H.

tutor
Arturo,
 
Yeah, I agree it's likely a typo (or a poorly written problem).
 
And you are very welcome!
 
Tony
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12/22/17

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Tony H. answered • 12/21/17

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The problem can be solved as written using Newton's 2nd Law:
 
Rotate the x- and y-axes so that the x-axis is parallel to the plane.  Use Newton's 2nd Law to sum the forces in the x-direction.  Since we are looking for a constant velocity, acceleration equals 0.  Find the x-components of the pulling force (F) and the gravity force (Fg) using trigonometry.
 
Fx - Fgx = ma =0
 
(F)(cos72) - (120N)(sin32) = 0
 
(F)(cos72) = (120N)(sin32)
 
F = ((120N)(sin32))/(cos72)
 
F = 205.8 N
 
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Arturo O.

Tony,
 
I obtained the same answer as you the first time I worked the problem.  But I also tried to find the normal force, and got a result that did not make sense to me (which is why I deleted my solution).  For a pull of 205.8 N and a normal force Fn, equilibrium of forces in the direction perpendicular to the plane gives
 
Fn + (205.8 N)sin72° = (120 N)cos32° ⇒
 
Fn = (120 N)cos32° - (205.8 N)sin72° ≅ -93.96 N
 
How can the normal force be negative?  What is wrong with this analysis?  If you can help me understand this, I will appreciate it!
 
 
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12/22/17

Arturo O. answered • 12/21/17

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Edited solution to be posted in Comment window.

 
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Mariana A.

Thank You!
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12/21/17

Arturo O.

Could you double-check the angles in the original problem statement?  With the angles as given, the applied force actually has a component down the plane instead of up the plane, and it would pull the block down the plane rather than up the plane.  The formula I used in my first solution is valid if the applied force makes an angle with the horizontal of less than 90°, but it makes an angle of (32+72)° = 104° with the horizontal, so it is pulling the block down, not up.  I realized that after I entered the solution.  This needs to be reworked with more realistic numbers.  If the ramp is inclined 32° and the force is pointing is 72° above the surface of the ramp, the force will pull the block down, not up.  I will rework the problem as soon as I see new numbers posted.
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12/21/17

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