Pulley but the cable is at an angle?

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  • #2
ehild
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Check your equation for the sliding box. You assume that the tension is equal to the weight of the hanging box. Think: If the tension was 280 N would the hanging box accelerate at all?
Check your signs in the equation for the vertical motion. Will the box accelerate upward or downward?

ehild
 
  • #3
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Check your equation for the sliding box. You assume that the tension is equal to the weight of the hanging box. Think: If the tension was 280 N would the hanging box accelerate at all?
Check your signs in the equation for the vertical motion. Will the box accelerate upward or downward?

ehild

I meant to say that the tension of the falling box is equal to the resultant tension of the sliding box. The tension in the x direction would be cos (10) * Tension of falling box.
 
  • #4
ehild
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Why did you took the tension 2.80 N?

ehild
 
  • #5
gneill
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Note also that as the block on the table moves towards the pulley, the angle of the cable will change (grow larger). The acceleration is going to change with the position of that block. Looks like there will be a differential equation to solve.
 
  • #6
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The tension (resultant and tension for the hanging box) was calculated using the weight of the hanging box.

Also, I'm in a Trig based physics course and we CAN'T do derivatives, actually we're not supposed to.
 
  • #7
gneill
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The tension (resultant and tension for the hanging box) was calculated using the weight of the hanging box.

Also, I'm in a Trig based physics course and we CAN'T do derivatives, actually we're not supposed to.

Then it would appear that the problem was poorly chosen for the class!

Note that the tension in the cable is also going to depend upon the acceleration, not just on the weight of the hanging block.

Suppose the cable were cut so that the hanging block could fall freely (equivalent to reducing the sliding mass to zero). What would be the tension in the cable? It would be zero because there's nothing to pull against! So clearly the tension depends upon both the masses and the acceleration of both.
 
  • #9
ehild
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Gneill is right, this problem is much harder I thought at first sight: It needs Calculus to solve.
If you get similar problems (but not complicated by angles) the tension in the rope is equal to the weight of the hanging mass only if it does not accelerate, as the acceleration of the hanging mass is a =(W-T)/m.


ehild
 
  • #10
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Gneill is right, this problem is much harder I thought at first sight: It needs Calculus to solve.
If you get similar problems (but not complicated by angles) the tension in the rope is equal to the weight of the hanging mass only if it does not accelerate, as the acceleration of the hanging mass is a =(W-T)/m.


ehild

I wish my PhD prof would realize this
 
  • #11
ehild
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What was the original text of the problem? Was the string said unstretchable? As the picture shows, the horizontal and vertical displacements can not be equal (the hypotenuse of the upper triangle is longer than its horizontal side).

ehild
 
  • #12
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In the diagram above, m1 = 285g, m2 = 755g, theta = 10.0 degrees, and mu = .047.
If system is released from rest, how long does it take m2 to reach the end of table.

Assum m1 can fall atleast 2.50 m.
 
  • #13
ehild
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OK, m2 can fall more than 2.5 m. But the problem is horrible. Ask your Prof how would he solve it. That angle will change unless the pulley is very far from the edge of the table.

ehild
 

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