Pulley Problem with External Force?

In summary, the conversation discusses a physics problem involving forces acting on a block and how to calculate the magnitude of one of the forces. There is a discrepancy in the given equation in the figure, but it is determined that the correct equation is 11g=(P+12g)μs. The conversation then goes on to solve for P using the equations P + Wb = Fn and Ff = fs*Fn. The resulting value for P is 251.9 N.
  • #1
Lori

Homework Statement



upload_2017-11-5_0-38-12.png

Homework Equations



Seen above

The Attempt at a Solution


Why is the force acting on block B = N + P + W =0? ?
Shouldn't it be B = -P-W+N?
So that P = -W+N=0
 

Attachments

  • upload_2017-11-5_0-38-12.png
    upload_2017-11-5_0-38-12.png
    63.4 KB · Views: 1,214
Physics news on Phys.org
  • #2
Note that the equation in the figure is a vector equation:
upload_2017-11-4_23-56-59.png


The three force vectors add to zero. But you are also correct that the magnitudes of these forces obey -P-W+N = 0.
 

Attachments

  • upload_2017-11-4_23-56-59.png
    upload_2017-11-4_23-56-59.png
    6.2 KB · Views: 867
  • #3
TSny said:
Note that the equation in the figure is a vector equation:
View attachment 214382

The three force vectors add to zero. But you are also correct that the magnitudes of these forces obey -P-W+N = 0.
I realized after that! But, I'm still confused on how they got that P =150?
 
  • #4
Lori said:
I realized after that! But, I'm still confused on how they got that P =150?
The μs factor was omitted. It should read 11g=(P+12g)μs.
 
  • #5
haruspex said:
The μs factor was omitted. It should read 11g=(P+12g)μs.
I'm still not sure how they calculated P=150 :/ Would appreciate it if anyone explained this one to me because I literally don't know how they added up and got 150. I tried multiplying the static friction too

Nevermind, i worked out the problem and figured thatP + W = Fn
Fƒ= μs×Fn
T=ƒ
T=Wa = 107.8=ƒ

Fn = Fƒ/Fƒs = 107.8/0.4 = 269.5
P = 269.5 - Wb = 269.5-117.6=251.9 N
 
Last edited by a moderator:
  • #6
Lori said:
I'm still not sure how they calculated P=150 :/ Would appreciate it if anyone explained this one to me because I literally don't know how they added up and got 150. I tried multiplying the static friction too
As haruspex noted, the equation 11g = P + 12g as given in the figure in post #1 is not correct; rather it should be 11g=(P+12g)μs.

Are you having trouble seeing where this equation comes from, or are you having trouble seeing how to solve this equation for P?
 
  • #7
TSny said:
As haruspex noted, the equation 11g = P + 12g as given in the figure in post #1 is not correct; rather it should be 11g=(P+12g)μs.

Are you having trouble seeing where this equation comes from, or are you having trouble seeing how to solve this equation for P?
Yes, i had trouble with where the equation was coming from, so i just redid it and found the answer myself! let me know if i did it correct cause my physics isn't really strong , so sometimes i doubt myself.!
 
  • #8
Lori said:
Nevermind, i worked out the problem and figured thatP + Wb = Fn
Ff = fs*Fn
T=f
T=Wa = 107.8=f

Fn = Ff/fs = 107.8/0.4 = 269.5
P = 269.5 - Wb = 269.5-117.6=251.9 N
Yes, that's it. Good.

(You use fs for μs. You can use the formatting toolbar to access Greek letters, subscripts, etc. Click on the symbol Σ at then end of the toolbar to access Greek letters and other math symbols.)
upload_2017-11-5_11-3-26.png
 

Attachments

  • upload_2017-11-5_11-3-26.png
    upload_2017-11-5_11-3-26.png
    1.7 KB · Views: 756
  • Like
Likes Lori
  • #9
TSny said:
Yes, that's it. Good.

(You use fs for μs. You can use the formatting toolbar to access Greek letters, subscripts, etc. Click on the symbol Σ at then end of the toolbar to access Greek letters and other math symbols.)
View attachment 214402
Thanks! I fixed it! I was actually wondering how to do that XD
 

Related to Pulley Problem with External Force?

What is a pulley problem with external force?

A pulley problem with external force is a type of physics problem that involves a system of pulleys and an external force acting on the system. The goal of these problems is to determine the tension and/or motion of the objects involved in the system.

What are the key components of a pulley problem with external force?

The key components of a pulley problem with external force are the pulleys themselves, the objects attached to the pulleys, and the external force acting on the system. These components are used to calculate the tension and motion of the objects in the system.

How do you approach solving a pulley problem with external force?

To solve a pulley problem with external force, you first need to identify all of the forces acting on the objects in the system. Then, you can use Newton's laws of motion and the principles of static equilibrium to set up and solve equations to determine the tension and motion of the objects.

What are some common mistakes when solving a pulley problem with external force?

One common mistake when solving a pulley problem with external force is not considering all of the forces acting on the system, including the weight of the pulleys themselves. Another mistake is not correctly setting up and balancing the equations, which can lead to incorrect solutions.

What are some real-world applications of pulley problems with external force?

Pulley problems with external force have many real-world applications, such as in engineering and construction projects that involve lifting and moving heavy objects. They are also used in the design of pulley systems for elevators and cranes, and in the study of mechanics and physics.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
729
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
468
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
1K
Back
Top