Magnitudes of the largest and smallest force?

  • Thread starter Fittler
  • Start date
  • #1
13
0
This is what I have and I have a feeling that I do not understand.

11. A box with a mass of 22 kg is at rest on a ramp inclined at 45 (degrees) to the horizontal. The coefficients of friction between the box and the ramp are: (mu(s): o.78 and mu(k): o.65)

a) Determine the magnitude of the largest force that can be applied upward parallel to the ramp, if the box is to remain at rest.

Fa = Fgsin45(degrees) - mu(k)Fncos45(degrees)
=53.4N

b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

Fa= Fgsin45(degrees) + mu(s)Fncos45(degrees)
=271.324N

So tell me, where di I go wrong?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,017
1,291
Fittler said:
This is what I have and I have a feeling that I do not understand.

11. A box with a mass of 22 kg is at rest on a ramp inclined at 45 (degrees) to the horizontal. The coefficients of friction between the box and the ramp are: (mu(s): o.78 and mu(k): o.65)

a) Determine the magnitude of the largest force that can be applied upward parallel to the ramp, if the box is to remain at rest.

Fa = Fgsin45(degrees) - mu(k)Fncos45(degrees)
=53.4N
Answer these questions:
(1) What's the normal force?
(2) What forces act on the box parallel to the ramp, including the applied force?
(3) What's the direction of the friction force?
(4) Since the box remains at rest, what must the net force be?

b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

Fa= Fgsin45(degrees) + mu(s)Fncos45(degrees)
=271.324N
Answer the same questions, only this time realize that the applied force is perpendicular to the ramp.

Also: Since the box remains at rest, only static friction is involved.
 
  • #3
13
0
So I drew a diagram and I am come up with the idea of needing touse the cosine. As in Cos45(deg)=Fn/9.8, which in turn gave me the force, and in turn I was able to use the Fs=(mu(s))Fn formula. Whatcha think? But I am still kind of stuck on the other one (b).
 
  • #4
Doc Al
Mentor
45,017
1,291
Not exactly sure what you are talking about. Why not answer the questions I posed? Step by step.
 
  • #5
13
0
Allright so, we will go through it together?

(1) What's the normal force?

The normal force is the force of gravity 9.8m/s^2 x mass
Since I do not know mass, it is 9.8m/s^2m

(2) What forces act on the box parallel to the ramp, including the applied force?

The forces that act parallel to the ramp on the box are the force of gravity, force of friction and the applied force.


(3) What's the direction of the friction force?

The friction force is upward, (parallel to the ramp) but in the case of pushing down on the box, the friction force would be downwards.

(4) Since the box remains at rest, what must the net force be?

F=ma or F=mg?

I dunno.
 
  • #6
Doc Al
Mentor
45,017
1,291
Fittler said:
Allright so, we will go through it together?
Good.

(1) What's the normal force?

The normal force is the force of gravity 9.8m/s^2 x mass
Since I do not know mass, it is 9.8m/s^2m
No, the normal force is not mg. That's the weight! (Yes, you are given the mass.)

(2) What forces act on the box parallel to the ramp, including the applied force?

The forces that act parallel to the ramp on the box are the force of gravity, force of friction and the applied force.
Good.


(3) What's the direction of the friction force?

The friction force is upward, (parallel to the ramp) but in the case of pushing down on the box, the friction force would be downwards.
I think you have it backwards. The friction force opposes slipping. When pushing up with max force, the friction pushes down; in the second problem, friction opposes gravity and thus acts upward.

(4) Since the box remains at rest, what must the net force be?

F=ma or F=mg?
I want a number. (Think equilibrium.)
 
  • #7
13
0
Allright:)

1. What's the normal force?

Is the normal force the cos of 45(deg) x 9.8? haha. I have no idea?

3. What's the direction of the friction force?

The friction force opposes motion of the applied force? So the friction force is acting downwards (parallel to the ramp). And the force of friction in part b) is acting to push the box upward, as a downward force is applied?

4. Since the box remains at rest, what must the net force be?

Thinking equillibrium, I could say that the normal force = the force of gravity?
 
  • #8
Doc Al
Mentor
45,017
1,291
Fittler said:
1. What's the normal force?

Is the normal force the cos of 45(deg) x 9.8? haha. I have no idea?
I'll give you this one. In part (a) the normal force is [itex]mg\cos\theta[/itex]. Not so for part (b).

3. What's the direction of the friction force?

The friction force opposes motion of the applied force? So the friction force is acting downwards (parallel to the ramp). And the force of friction in part b) is acting to push the box upward, as a downward force is applied?
Correct. (Mostly.)

4. Since the box remains at rest, what must the net force be?

Thinking equillibrium, I could say that the normal force = the force of gravity?
What's the net force parallel to the ramp?
 
  • #9
I'll give you this one. In part (a) the normal force is [itex]mg\cos\theta[/itex]. Not so for part (b).


Correct. (Mostly.)


What's the net force parallel to the ramp?
hi again!

ok so i was looking at this problem and well i kind of came to this conclusion from the last quote on here.

Fnet=Fn+mgy
=mg+mgcostheta
=(22kg)(9.8m/s2)+(22kg)(9.8m/s2)cos45
=215.6+152.45
=367.45

the net force parallel to the ramp.
is that the net force? am I on the right track?
 
  • #10
ok, i'm not sure if my last post was anywhere near correct...

I reworked it and this is what i came up with...

parallel forces to the ramp
Fg, Fs, Fa

sum of F= Fa - Fs - Fg
Fa=Fs + Fg
=usFn+mg
=(us)(mg)costheta+mg <-- do the mg cancel out?
=(us)costheta <--- I cancelled them out
=(7.8)cos45
=5.5N
 
  • #11
Doc Al
Mentor
45,017
1,291
I'm not quite sure what you're doing. For one thing, the weight of the box acts vertically, not parallel to the incline.

Try answering the questions that I posed in post #2.
 
  • #12
ok so after good session of thinking... lol

here are the answers to the questions you posed in post #2
(1) What's the normal force?

the normal Force is Fn=FgNcostheta

(2) What forces act on the box parallel to the ramp, including the applied force?

Fs,Fa,Fgr

(3) What's the direction of the friction force?

downward

(4) Since the box remains at rest, what must the net force be?

if it will remain at rest both forces must be equal
so, Fa=Fs
 
  • #13
Doc Al
Mentor
45,017
1,291
here are the answers to the questions you posed in post #2
(1) What's the normal force?

the normal Force is Fn=FgNcostheta
Are you multiplying the force of gravity (Fg) by the normal force (N)?
(2) What forces act on the box parallel to the ramp, including the applied force?

Fs,Fa,Fgr
Which direction does Fg act? What component is parallel to the ramp?
(3) What's the direction of the friction force?

downward
OK. Downward and parallel to the ramp.
(4) Since the box remains at rest, what must the net force be?

if it will remain at rest both forces must be equal
so, Fa=Fs
What happened to gravity?
 
  • #14
ummm... ok i'm really really confused...

do you mind solving the problem so i can understand how to even start it?
 
  • #15
This post is for question b) The box is being pushed directly against the ramp. Therefore, why would the box move?

Wouldn't the smallest force that can be applied to the top of the box without it moving be 0.

Or are they asking about the smallest force that would make the box move? I don't know.

Is the box, like the question says, being pushed into the ramp or down it?
 
Last edited:
  • #16
Doc Al
Mentor
45,017
1,291
In question b) the applied force is perpendicular to the ramp (pushing the block directly against the ramp). If that force were zero, would the block slide down the ramp? Find out by analyzing the forces acting on the block. (Other forces act on the block, not just the applied force.)
 
  • #17
thank you so much. your post just made me realize what i have to do.
 
  • #18
So, what you are saying is that it is as if we can control the force of gravity, and we need to find out the minimum force of gravity needed to keep this box from sliding. Am I correct?
 
  • #19
Doc Al
Mentor
45,017
1,291
So, what you are saying is that it is as if we can control the force of gravity, and we need to find out the minimum force of gravity needed to keep this box from sliding. Am I correct?
No. When the box is pushed directly into the ramp, gravity doesn't change. But the normal force does, which changes the friction force. Find the minimum friction force needed to keep the box from sliding. Then figure out the added force required to create such a friction force.
 
  • #21
Hold on, maybe I don't see. I did the question and I got a value that was higher than the original normal force, which is impossible seeing how the original normal force held the box in rest.
 
  • #22
Doc Al
Mentor
45,017
1,291
Hold on, maybe I don't see. I did the question and I got a value that was higher than the original normal force, which is impossible seeing how the original normal force held the box in rest.
How did you reach that conclusion?
 
  • #23
Quite simple. The question states that the box is at rest. And when the normal force is calculated, it is smaller than the minimum normal force needed to hold the box in rest.
 
  • #24
Doc Al
Mentor
45,017
1,291
Quite simple. The question states that the box is at rest.
You might want to reread the problem statement:
"b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest."​
Nothing in there says that the box would remain at rest if you didn't apply some kind of force.
And when the normal force is calculated, it is smaller than the minimum normal force needed to hold the box in rest.
So what does that tell you?
 
  • #25
Whoa, I am a slow thinker. Sorry about that, I should have seen it a long time ago. The needed force would be that larger number I had minus the normal force. That I should have seen. Sorry and thanks for the help!
 

Related Threads on Magnitudes of the largest and smallest force?

Replies
3
Views
5K
Replies
5
Views
2K
Replies
2
Views
17K
Replies
2
Views
1K
  • Last Post
Replies
10
Views
7K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
Replies
2
Views
7K
Replies
2
Views
6K
Replies
12
Views
2K
Top