Pulley but the cable is at an angle?

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a pulley system where the cable is at an angle. Participants are tasked with finding the acceleration of a heavier box to determine how long it takes to reach the end of a table.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between tension and weight in the context of the pulley system, questioning assumptions about acceleration and the effects of the angle of the cable. Some suggest that the problem may require calculus due to the changing angle as the block moves.

Discussion Status

The discussion is ongoing, with participants offering insights and raising questions about the setup and assumptions of the problem. There is recognition that the problem may be more complex than initially thought, and some participants express concerns about the appropriateness of the problem for the course level.

Contextual Notes

Participants note constraints such as the prohibition of derivatives in their course and the potential implications of the angle changing as the block moves. There is also mention of the need for clarification on the original problem statement regarding the properties of the string.

Physics news on Phys.org
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
 
ehild said:
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.
 
Why did you took the tension 2.80 N?

ehild
 
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.
 
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.
 
amd123 said:
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.
 
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
ehild said:
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
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
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
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
 

Similar threads

Solving for the unknown variable in Motional EMF problem | Physics Homework
  • · Replies 7 ·
Replies
7
Views
2K
Calculating Repulsive Force Between Parallel Wires
  • · Replies 2 ·
Replies
2
Views
2K
Defining equilibrium and movement in pulley arrangements
  • · Replies 8 ·
Replies
8
Views
3K
How Do You Calculate Cable Length and Vector Forces in Statics?
  • · Replies 15 ·
Replies
15
Views
5K
The Relationship Between Masses and Angles in a Pulley System
  • · Replies 29 ·
Replies
29
Views
4K
What is the angular acceleration of the rod?
  • · Replies 18 ·
Replies
18
Views
7K
Proving T=H+q*h in a Catenary Spanned Cable
  • · Replies 7 ·
Replies
7
Views
2K
Motor lowers a mass over a pulley, find acceleration of mass
  • · Replies 3 ·
Replies
3
Views
2K
What is the Electric Potential of a Square Loop?
  • · Replies 2 ·
Replies
2
Views
7K
Equilibrium : normal reaction of a rotational center?
  • · Replies 2 ·
Replies
2
Views
2K