What is the minimum tension (in Newtons) in the angled rope?

  • Thread starter bah83
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  • #1
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Homework Statement


A hanging 2.00-kg (19.6N) block begins to accelerate a 4.28-kg (41.94N) block on a horizontal surface. But a rope connected to the wall at 26.9 degrees brings the block to an equilibrium position. The coefficient of friction between the surface and the 4.28-kg (41.94N) block is 0.162.

What is the minimum tension (in Newtons) in the angled rope that maintains this equilibrium condition?

The answer is 15.6 Newtons but I cannot figure out how to get there.
Here's a picture: http://gbhsweb.glenbrook225.org/gbs/science/phys/chemphys/audhelp/u8setf/q14.gif

Homework Equations


T-mgsinθ=ma so T=ma+mgsinθ
F=ma

The Attempt at a Solution


I've tried many different ways. I've tried to find both the horizontal and vertical forces of the rope and picturing it as a triangle. I may just be adding the forces wrong or using the wrong equations. I found that the frictional force of the 41.94N block is 6.79N. I thought that the tension of the rope between the two blocks is 19.6N (equal to the weight of the hanging one). I thought I had found the answer when I subtracted 6.79+19.6 from 41.94 (equals 15.55). But I know that isn't the right way to do it. Help please.
 

Answers and Replies

  • #2
Simon Bridge
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Show us your working one step at a time ... do not substitute any numbers until all the algebra is finished.

Most likely mistakes include:
getting the trig wrong for the rope to the wall.
forgetting the upwards pull of that rope when calculating friction.
 
  • #3
arildno
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Other possible source errors are trivial arithmetical mistakes
Recently, one student here had made a very long and good calculation, but made the mistake 16*2=28 at one point.
Such mistakes occur OFTEN, for all of us, and is one of the major reasons why you should MINIMIZE the number of steps in which you use numbers, rather than letters.
 

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