Need help in the physic HW question This is not the box question

  • Thread starter NasuSama
  • Start date
  • Tags
    Box Physic
In summary: So, the mgcosθ must be tilted so that it points downward along the handle in order to account for this.
  • #1
NasuSama
326
3

Homework Statement



The handle of a floor mop makes an angle θ = 28.6o with the horizontal. Assume the handle is massless, and the mop has mass M = 1.4 kg. The coefficient of kinetic friction between the mop and the floor is μk = 0.386. Find F, the magnitude of the force, exerted downward along the handle, that will cause the mop to slide across the floor without acceleration.

Homework Equations



→ Σ F_up = Σ F_down and Σ F_left = Σ F_right
→ Some trig. functions like the one with cosine and sine.
→ F_r = µF_N

The Attempt at a Solution



I drew the free body diagram, and I have mgsin(θ) in the "down" direction. Also, since you are exerting the force downward, F must be also in "down" direction. I believe that the normal force is F_N "up". Then, the friction force goes "left" direction, and the mgcos(θ) goes "right direction". I also drew the mop forming 28.6° with the horizontal.

I am thinking of forming these equations...

F_N = F + mgsin(θ) [up = down]
F_r = mgcos(θ) [left = right]

..Then, I solve for F and got 24.7 N, which is the wrong answer.
 
Physics news on Phys.org
  • #2
By "up or down" I assume you mean vertical (perpendicular to the floor). Now, the force F does not act perpendicular to the floor, so you'll need a trig function to get the vertical component of F. Why multiply mg by sinθ to get the vertical component? What direction does mg point?
 
  • #3
TSny said:
By "up or down" I assume you mean vertical (perpendicular to the floor). Now, the force F does not act perpendicular to the floor, so you'll need a trig function to get the vertical component of F. Why multiply mg by sinθ to get the vertical component? What direction does mg point?

Yes, that is what I mean by "up or down". I'm referring to vertical components.

Well, if you exert the mop downward, then that would be F/cosθ for F. Am I right?

Oh! My bad. mg always points down. Let me know if I'm in the right track.

This is what I leave off...

F/cos(θ) + mg = F_N
F_r = mgcos(θ)
 
  • #4
NasuSama said:
Well, if you exert the mop downward, then that would be F/cosθ for F. Am I right?

No, that's not it. Did you construct a diagram showing the direction of F relative to horizontal? Can you sketch the horizontal and vertical components of F on the diagram?
 
  • #5
TSny said:
No, that's not it. Did you construct a diagram showing the direction of F relative to horizontal? Can you sketch the horizontal and vertical components of F on the diagram?

Still no clue. Here is the diagram I drew. Check attachment.
 

Attachments

  • help.JPG
    help.JPG
    11.7 KB · Views: 326
  • #6
Does mg have a horizontal (left, right) component?

You've drawn a force straight down and labeled it F. What does the problem tell you about the direction of F?

What is the force that you've drawn up and to the left at the angle θ? It doesn't have a label.
 
  • #7
TSny said:
Does mg have a horizontal (left, right) component?

You've drawn a force straight down and labeled it F. What does the problem tell you about the direction of F?

What is the force that you've drawn up and to the left at the angle θ? It doesn't have a label.

Sorry about that. That is labelled the mop, and F is shown to be exerted downward. If F is exerted downward at that mop, then would that be Fsin(θ)?
 
  • #8
F is given to be in a direction "downward along the handle". That means that the direction of F is along the handle toward the mop head. It's in the same direction as the handle.

I don't understand the force that you call "mop". Your diagram should only include forces acting on the mop head. Where does the force you call "mop" come from?
 
  • #9
TSny said:
F is given to be in a direction "downward along the handle". That means that the direction of F is along the handle toward the mop head. It's in the same direction as the handle.

I don't understand the force that you call "mop". Your diagram should only include forces acting on the mop head. Where does the force you call "mop" come from?

I posted the revised diagram. This is the mop I am demonstrating.
 

Attachments

  • help.JPG
    help.JPG
    14 KB · Views: 326
  • #10
You need to tilt the force F so that it points down along the mop handle.

Also, think about the mgcosθ. If mg points downward, then how much of mg points horizontally?
 
  • #11
The interpretation of the wording of the problem is causing some confusion. The question reads, "Find F, the magnitude of the force, exerted downward along the handle." The word "downward" here does not mean vertically downward. F points along the handle - down and to the right at angle θ relative to the horizontal.
 
  • #12
TSny said:
The interpretation of the wording of the problem is causing some confusion. The question reads, "Find F, the magnitude of the force, exerted downward along the handle." The word "downward" here does not mean vertically downward. F points along the handle - down and to the right at angle θ relative to the horizontal.

Then, this means that I need to find the force perpendicular to the handle of the mop? Is that what you mean?
 
  • #13
The force, F, that you are looking for is the force that the mop handle pushes on the head of the mop. You are given that the direction of this force is along the handle; that is, F makes the same angle with respect to the horizontal as the handle does. So, F slopes down and to the right making an angle of 28.6o below the horizontal. Your free body diagram for the head of the mop should show this force along with the weight, normal force, and friction force.
 
  • #14
TSny said:
The force, F, that you are looking for is the force that the mop handle pushes on the head of the mop. You are given that the direction of this force is along the handle; that is, F makes the same angle with respect to the horizontal as the handle does. So, F slopes down and to the right making an angle of 28.6o below the horizontal. Your free body diagram for the head of the mop should show this force along with the weight, normal force, and friction force.

Then, the force pushes in the way that the mop is held 28.6° with the horizontal, yes? I am totally confused. Sorry to get bumped so easily.
 
  • #15
The attached picture shows the direction of the force you're trying to determine.
 

Attachments

  • Force F.jpg
    Force F.jpg
    5.4 KB · Views: 391
  • #16
TSny said:
The attached picture shows the direction of the force you're trying to determine.

Hm... So this could show that F = mgsin(θ) for the vertical component and I can use it for the equation set up. Hopefully, I'm on the right track.
 
  • #17
How did you get F = mgsinθ?
 
  • #18
TSny said:
How did you get F = mgsinθ?

I don't know. It's basically that I use the trigonometry form. mg, which I am applying on the mop, is the hypotenuse. Then..

Oh never mind. I am not sure about the next steps. I am wrong about this.
 
  • #19
You have the right idea of setting up Σ F_up = Σ F_down and Σ F_left = Σ F_right. You just need to be very clear on the direction of the forces (so, a good free-body diagram) and you need to be able to find the horizontal and vertical components of the forces.
 
  • #20
O.K. I should get the correct answer. Let's see. Based on the attached picture you showed to me and the formulas I use, I obtain:

F_N = mg + Fsin(θ) [Σ F_up = Σ F_down]
F_r = µ_k * F_N = Fcos(θ) [Σ F_left = Σ F_right]

µ_k(mg + Fsin(θ)) = Fcos(θ)
µ_kmg + Fµ_ksin(θ) = Fcos(θ)
µ_kmg = Fcos(θ) - Fµ_ksin(θ)
F = µ_kmg/(cos(θ) - µ_ksin(θ))

That is the force without the value substitutions. Let's see if I'm really on the right track this time! :D
 
  • #21
Very good! Nice work. :smile:
 
  • #22
Thanks for the help! I got the right answer, and it's right!

I will ask another problem in the different thread. It's different from this problem.
 

1. What is the question asking for?

The question is asking for assistance in solving a specific physics homework problem. It is not related to the common "box question" often used in physics textbooks.

2. Can you provide any hints or tips for solving the problem?

Without knowing the specific details of the problem, it is difficult to provide specific hints or tips. However, a good approach is to carefully read and understand the question, identify the relevant equations and variables, and then apply the appropriate equations to solve the problem.

3. Do you have any recommended resources for further understanding the topic?

There are many online resources available for learning and understanding physics concepts. Some recommended resources include Khan Academy, Physics Classroom, and HyperPhysics. Additionally, consulting your textbook or asking your teacher for clarification can also be helpful.

4. What should I do if I am still struggling with the problem?

If you are still struggling with the problem, it is important to not get discouraged. Take a break and come back to the problem later with a fresh perspective. You can also try working with a study group or seeking assistance from a tutor or your teacher.

5. How can I check if my answer is correct?

You can check your answer by comparing it to the answer provided in the textbook or by double-checking your calculations and units. If you are still unsure, you can ask your teacher or a classmate for their input. It is also helpful to understand the context of the problem and whether your answer makes logical sense.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
9K
Replies
5
Views
7K
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
18
Views
3K
Back
Top