Pulley at an Angle Problem - Newton Forces

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SUMMARY

The discussion centers on solving the Pulley at an Angle problem using Newton's laws of motion. Participants analyze the forces acting on two blocks connected by a rope, focusing on tension, normal force, and friction. Key equations include Fnet = ma and the friction force equation, which is derived from the normal force multiplied by the static coefficient of friction. The consensus is that the tension in the rope equals the weight of the hanging block, and participants emphasize the importance of writing out equilibrium equations for both the x and y directions to arrive at the correct solution.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free-body diagrams
  • Familiarity with static friction and its calculation
  • Ability to solve equilibrium equations in two dimensions
NEXT STEPS
  • Study the derivation of the equations of motion for pulleys
  • Learn about static and kinetic friction coefficients
  • Explore advanced topics in dynamics, such as torque and rotational motion
  • Practice solving similar problems involving multiple bodies and forces
USEFUL FOR

Students studying physics, particularly those tackling mechanics and dynamics problems involving pulleys and forces. This discussion is beneficial for anyone looking to deepen their understanding of equilibrium and tension in systems with multiple objects.

Lori

Homework Statement


upload_2017-11-7_20-18-20.png


Homework Equations


Friction Force = normal force * static coefficient for friction
Fnet = ma
upload_2017-11-7_20-20-17.png


The Attempt at a Solution



So i drew the free-body diagrams for both the objects...[/B]
Im trying to figure out the tension in the 2nd free body diagram and i know that it's equal to the weight of the object (which i want)...

I was wondering which tension in the first body diagram equals the tension to the 2nd object? I found that Friction = 684 (1520*0.45) thus the component force in the x direction for the rope should be 684.

I then found the hypotenuse of that to find Tension, which is Tcos(60) = 684 and found that T = 1184... but it's not really that close to the answers in the problem! (it is closest to 1898 N, but I am not sure if that's the correct answer)

What did i do wrong!?
 

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Can you show the equation that you used to get the normal force and from it the maximum force of static friction?
 
I'm not sure what it is you're doing. Why are you trying to find tension? As you said, the tension is the weight of the hanging block, so you already the know tension. The problem is to then find the maximum weight of the hanging block (i.e., maximum tension) for which the system is stationary.

Under what condition will the first block begin to move in the x direction?
 
RedDelicious said:
I'm not sure what it is you're doing. Why are you trying to find tension? As you said, the tension is the weight of the hanging block, so you already the know tension. The problem is to then find the maximum weight of the hanging block (i.e., maximum tension) for which the system is stationary.

Under what condition will the first block begin to move in the x direction?
Wait I'm confused. I don't know tension cause I don't know the weight of the hanging block?
 
You don't know what it is but you know it is equal to the hanging weight because everything is at equilibrium. The hanging weight is what you are looking for, so call it W and proceed with the calculation. In the end you should be able to find a value for W.
 
Lori said:
Wait I'm confused. I don't know tension cause I don't know the weight of the hanging block?
RedDelicious' post is rather confusing. Not sure what was intended. Ignore that and try to answer Kuruman's question in post #2.
Lori said:
was wondering which tension in the first body diagram equals the tension to the 2nd object?
There is only one tension in each diagram. Can you think of a reason why they should or should not be equal?
 
haruspex said:
RedDelicious' post is rather confusing. Not sure what was intended. Ignore that and try to answer Kuruman's question in post #2.

There is only one tension in each diagram. Can you think of a reason why they should or should not be equal?
Hold on, I'll work on this problem later, but to answer your question , I suppose the rope that connect the objects must have the same tension
 
Lori said:
the rope that connect the objects must have the same tension
Yes, but it is interesting to prove it. Consider the torque balance on the pulley.
 
haruspex said:
RedDelicious' post is rather confusing. Not sure what was intended. Ignore that and try to answer Kuruman's question in post #2.

There is only one tension in each diagram. Can you think of a reason why they should or should not be equal?
I'm trying to solve for W , but i keep getting that T = 1368 = W??

I did this since Friction force would be 684 so that means that x components must balance out for the x forces. Thus, i did the 684/cos(60) to find the tension of the rope which i know equals the weight of the hanging mass??
 
  • #10
Lori said:
since Friction force would be 684
That is why you need to respond to kuruman's post #2. Write out the equations by which you deduce that.
 
  • #11
haruspex said:
That is why you need to respond to kuruman's post #2. Write out the equations by which you deduce that.

Friction = 0.4N

N = cos60 = ?/T

N + T = 1520

Friction = 0.4 (1520-T)
 
  • #12
Lori said:
Friction = 0.4N
Yes.
Lori said:
N = cos60 = ?/T
I don't understand what you mean by that.
Lori said:
N + T = 1520
The tension is not vertical.
 
  • #13
haruspex said:
Yes.

I don't understand what you mean by that.

The tension is not vertical.
Is normal force not going to be equal to the weight of the object because tension is involved? So that the tension's y component + the normal force = 1520 ( weight of object)
 
  • #14
Here is solution hope u understand and T=weight of hanging block
 
  • #15
Lori said:
Is normal force not going to be equal to the weight of the object because tension is involved? So that the tension's y component + the normal force = 1520 ( weight of object)
Right.
 
  • #16
Abhishek kumar said:
Here is solution hope u understand and T=weight of hanging block
This is a homework forum. The rule is to post hints, point out errors, etc. Please do not post complete solutions.
 
  • #17
Ahhh. Thank you. I understand what you guys were wanting me to do... I think the problem was that I plugged in numbers to early on, and I should have written the equations that showed equilibrium in the x and y directions. .

Thanks everyone [emoji32]

Slowly understanding physics!
 
  • #18
Ok
haruspex said:
This is a homework forum. The rule is to post hints, point out errors, etc. Please do not post complete solutions.
Ok from now onwards i will keep in mind that
 

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