Relationship between horizontal force and friction in an exercise

  • #1
Venturi365
12
3
Homework Statement
A mass of ##20\,\mathrm{N}## lays on a horizontal surface. The static friction coefficient and the kinetic friction coefficient are, respectively: ##\mu_{s}=0.8## and ##\mu_{k}=0.6##. A horizontal cable is tied to the mass with a constant tension ##T##. What's the friction force that acts on the block if ##T=15\,\mathrm{N}## and ##T=20\,\mathrm{N}##?
Relevant Equations
##f_{s}=\mu_{s}\cdot N##
##f_{k}=\mu_{k}\cdot N##
The thing with this exercise is that I don't think that the question makes sense at all (or, at least, is incomplete).

First of all, we don't know if the mass moves with any of those tensions, therefore I cannot know which coefficient apply. Second of all, even if we suppose that the mass is moving or not, the friction force has nothing to do with any horizontal force, but rather it is related to the movement and weight of the mass. Right?

Anyway, if we suppose that in both cases the mass doesn't move, then:

$$
\begin{align}
f_{s}=0.8\cdot 20 =16\,\mathrm{N}
\end{align}
$$

If we suppose that in both cases the mass moves, then:

$$
\begin{align}
f_{k}=0.6\cdot 20=12\,\mathrm{N}
\end{align}
$$

Am I missing something or is it the wording of the problem just uncomplete or misleading?
 
Physics news on Phys.org
  • #2
Venturi365 said:
Homework Statement: A mass of ##20\,\mathrm{N}## lays on a horizontal surface. The static friction coefficient and the kinetic friction coefficient are, respectively: ##\mu_{s}=0.8## and ##\mu_{k}=0.6##. A horizontal cable is tied to the mass with a constant tension ##T##. What's the friction force that acts on the block if ##T=15\,\mathrm{N}## and ##T=20\,\mathrm{N}##?
Relevant Equations: ##f_{s}=\mu_{s}\cdot N##
##f_{k}=\mu_{k}\cdot N##

The thing with this exercise is that I don't think that the question makes sense at all (or, at least, is incomplete).

First of all, we don't know if the mass moves with any of those tensions, therefore I cannot know which coefficient apply. Second of all, even if we suppose that the mass is moving or not, the friction force has nothing to do with any horizontal force, but rather it is related to the movement and weight of the mass. Right?

Anyway, if we suppose that in both cases the mass doesn't move, then:

$$
\begin{align}
f_{s}=0.8\cdot 20 =16\,\mathrm{N}
\end{align}
$$

If we suppose that in both cases the mass moves, then:

$$
\begin{align}
f_{k}=0.6\cdot 20=12\,\mathrm{N}
\end{align}
$$

Am I missing something or is it the wording of the problem just uncomplete or misleading?
Notice that the coefficient of static friction is larger than the kinetic. Initially the block is "laying there". In order to get it to move you must overcome the force of static friction. Once moving the frictional force coefficient sharply drops off to ##\mu_k##

Concept question: When the block is sitting there without any horizontal force applied. What is the force of friction?
 
  • #3
erobz said:
Notice that the coefficient of static friction is larger than the kinetic. Initially the block is "laying there". In order to get it to move you must overcome the force of static friction. Once moving the frictional force coefficient sharply drops off to ##\mu_k##

Ohhhh, so therefore when ##T=15## then ##f\leq 16## and when ##T=20## then ##f=12##
 
  • #4
Venturi365 said:
Ohhhh, so therefore when ##T=15## then ##f\leq 16##
No, you are on the right track but this statement is not correct.

When the block is sitting there without any horizontal tension force, what is the force of static friction acting on it?
 
  • #5
erobz said:
When the block is sitting there without any horizontal tension force, what is the force of static friction acting on it?

Well, since the friction is a reaction force, it has to be ##f=0\,\mathrm{N}##

I guess that technically, when ##T=15\,\mathrm{N}## then ##f=15\,\mathrm{N}## because there's no resulting movement and ##\mu_{s}=0.8## is just the maximum value that it could take.
 
  • Like
Likes erobz
  • #6
Venturi365 said:
Well, since the friction is a reaction force, it has to be ##f=0\,\mathrm{N}##

I guess that technically, when ##T=15\,\mathrm{N}## then ##f=15\,\mathrm{N}## because there's no resulting movement and ##\mu_{s}=0.8## is just the maximum value that it could take.
There you go!
 
  • Like
Likes Venturi365
  • #7
erobz said:
There you go!

Thank you for your help!
 
  • Like
Likes erobz
  • #8
Venturi365 said:
Relevant Equations: ##f_{s}=\mu_{s}\cdot N##
No. $$f_s\leq \mu_s~N##. This says that the force static friction is at most ##\mu_s~N##. Otherwise, it's whatever is necessary to provide the observed acceleration. It's like a credit card with a maximum of $2,000. When you go to a store to buy something, you can use the card to buy anything under $2,000. What you charge to the card is the price of the object, not $2,000.
 

Related to Relationship between horizontal force and friction in an exercise

What is the relationship between horizontal force and friction during exercise?

The relationship between horizontal force and friction during exercise is that friction opposes the horizontal force applied. When you apply a horizontal force to move an object or your body, friction acts in the opposite direction, affecting the efficiency and difficulty of the movement.

How does increasing horizontal force affect friction in an exercise?

Increasing the horizontal force generally increases the frictional force up to a certain point. This is because friction is proportional to the normal force, which is influenced by the horizontal force applied. However, once the maximum static friction is overcome, the object starts to move, and kinetic friction takes over, which is usually lower than static friction.

What role does friction play in exercises involving horizontal force?

Friction plays a crucial role in exercises involving horizontal force by providing the necessary resistance that muscles need to work against. It helps in stabilizing movements, preventing slips, and ensuring controlled motion. Without friction, it would be difficult to perform many exercises safely and effectively.

How can friction be reduced or increased during an exercise?

Friction can be reduced by using smoother surfaces, lubricants, or specialized equipment like sliders or resistance bands. It can be increased by using textured surfaces, adding weight, or using equipment with higher friction coefficients. Adjusting friction levels can help in tailoring the difficulty and effectiveness of an exercise.

Why is understanding the relationship between horizontal force and friction important in exercise science?

Understanding this relationship is important because it helps in designing effective and safe exercise programs. It allows for the optimization of resistance levels, improves performance, and minimizes the risk of injury. Knowledge of how friction interacts with horizontal force can also aid in the development of better exercise equipment and surfaces.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
417
  • Introductory Physics Homework Help
Replies
8
Views
493
  • Introductory Physics Homework Help
Replies
13
Views
680
  • Introductory Physics Homework Help
Replies
17
Views
875
  • Introductory Physics Homework Help
Replies
8
Views
789
  • Introductory Physics Homework Help
Replies
33
Views
333
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
225
  • Introductory Physics Homework Help
Replies
10
Views
829
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top