Mechanical Physics: Coefficient of Friction

[Daniel Guh]

Homework Statement

A force of 300 N is applied to a 20 kg crate at an angle of 30o below the horizontal. Determine the acceleration of the crate as it moves along the horizontal surface. The coefficient of friction between the crate and the surface µk is 0.50.
1. 17.2
2. 8.6 (My choice but it's incorrect)
3. 4.3
4. 9.8

Relevant Equations

Fg = mg
μ = Ff / Fn
F = ma

First of all, the pulling force is
300N cos(30) = 260 N
At this point, I try to find the friction force
Fn = mg = 20kg * 9.81 m/s^2 = 196.2 N
Then,
Ff = μ * Fn = 0.5 * 196.2 N = 98.1 N
So after canceling the horizontal forces,
260N - 98.1N = 161.9N
And the acceleration will be 161.9N / 20kg = 8.1 m/s ^2
Since 8.6 is the closest I choose 8.6, but it's incorrect.

 

[malawi_glenn]

The normal force is not mg in this case

 

[Daniel Guh]

The normal force is not mg in this case

Ohh, do I have to include the force from the 300N? Since there is a downward force, the normal force will be greater?

 

[malawi_glenn]

Since there is a downward force, the normal force will be greater?

Yes. Let's say you have a book lying on a table. If you push down on the book with your hand, the normal force from the table on the book will be greater than $m_\text{book}g$.

 

[MatinSAR]

@Daniel Guh Can you draw a free body diagram?
Note that $\vec F_{net}=m\vec a$ is equivalent to three component equations:
$F_{net,x}=ma_x$
$F_{net,y}=ma_y$
$F_{net,z}=ma_z$
If there is no motion in z direction it means that $a_z$ is $0$ so we should have $F_{net,z}=0$ .

 

[mathguy_1995]

Consider again the normal force. Part of the force $F$ and particularly its component $F\cdot \sin (30)$ participates into that computation. As said above, if someone pushes downwards a book (placed on a table), the normal force will not be equal to its weight, but greater, otherwise, the book could've not stood on it.

 

[Lnewqban]

Ohh, do I have to include the force from the 300N? Since there is a downward force, the normal force will be greater?

Welcome, @Daniel Guh !

May be I am wrong, but to me, the statement “A force of 300 N is applied to a 20 kg crate at an angle of 30o below the horizontal.” does not mean downward.

 

[kuruman]

Welcome, @Daniel Guh !

May be I am wrong, but to me, the statement “A force of 300 N is applied to a 20 kg crate at an angle of 30o below the horizontal.” does not mean downward.

To me, "above" and "below" the horizontal are synonymous with above the horizon and below the horizon. The sun rises above the horizon and sets below the horizon. The reference point is at the horizon. So if you have a vector below the horizontal, this conveys (at least to me) that the tail of the vector is at the horizontal and its tip below the horizontal. If this force vector were parallel to the position vector of the Sun relative to the horizon, it would describe the Sun at dusk.

 

[hmmm27]

May be I am wrong, but to me, the statement “A force of 300 N is applied to a 20 kg crate at an angle of 30o below the horizontal.” does not mean downward.

That's also how I read it ...however the answer for that is not in the list.

It would have been so easy to phrase it like "A force of 300N is applied downwards, 30deg from the horizontal" or summat.

 

[nasu]

Welcome, @Daniel Guh !

May be I am wrong, but to me, the statement “A force of 300 N is applied to a 20 kg crate at an angle of 30o below the horizontal.” does not mean downward.

Then, how would you understand this statement?

 

[haruspex]

That's also how I read it ...however the answer for that is not in the list.

Not sure which way you say you would read it. In my experience the terminology is fairly standard. It is below the horizontal from the perspective of whatever is applying the force, so tending downward
And it does give me an answer in the list.

 

[hmmm27]

And it does give me an answer in the list.

Whereas, treating it as an upward force does not... which is what I said.

But, the question does look unambiguously as if an upwards force is being applied, despite usage of "at" rather than "from" or "to".

If one were the applier of a downwards(ish) force, the obvious place to push would be the top of the crate and the obvious measurement for the angle would be from horizontal *upwards*. Why would you break open the crate and take the measurement there ?

 

[nasu]

Who said anything about breaking the box? You can push on top of the crate at any angle you want, above, below or along the horizontal.

 

[haruspex]

the obvious measurement for the angle would be from horizontal *upwards*

Because you are thinking of it in regards to drawing a FBD, maybe, not from the perspective of the person, or whatever, applying the force.
How would you aim a rod if told to point it below the horizontal?

 

[Lnewqban]

Then, how would you understand this statement?

If not downward, the only alternative is upward.
I see @kuruman point of view, considering tail of the vector attached to the box, or it always being pulled rather than pushed.
Again, I may be wrong, but in practical terms, one pushes a box down-forward and pulls it up-forward.
Sorry about creating more confusion, @Daniel Guh

 

[nasu]

The statement does not mention either pulling or pushing. The force "is applied".

 

[malawi_glenn]

Again, I may be wrong, but in practical terms, one pushes a box down-forward and pulls it up-forward.

Consider pushing vs pulling a car. Are these two options charactrized by how the vertical force component is directed? Or is it something else.

Btw the original problem formulation just said that a force was appllied. No pulling or pushing was mentioned

 

[haruspex]

in practical terms, one pushes a box down-forward and pulls it up-forward.

Only because you are assuming you are taller than the box.

 

[kuruman]

I see @kuruman point of view, considering tail of the vector attached to the box, or it always being pulled rather than pushed.

Forces in a FBD are assumed to act at the CoM and are drawn with their tails at the CoM to facilitate figuring out the angles needed to add them.

Whether a pushing as opposed to pulling force acts on the system is irrelevant to how one draws the arrow representing this force. Consider a block hanging from a sting and one being supported by a table. The appropriate force arrow is drawn the same and there is nothing to distinguish the "pulling" from the "pushing" force.

(Edited to add clarity)

 

[malawi_glenn]

Forces in a FBD are assumed to act at the CoM

Never seen that assumption before
https://en.wikipedia.org/wiki/Free_body_diagram

 

[p-waive]

The wording of the problem might be confusing a bit. However, if you try a few combinations of calculation you will see the result should be smaller than your calculated one. This diagram presents the quiz probably better:

This is probably clear already.