Maximum force applied to prevent block from sliding up ramp

In summary: So if you want the maximum force that won't cause the block to slide up, then you want F to be smaller than W. Why? Is [tex]1-\mu_s \tan\theta[/tex] always...yes, it is always negative. So if you want the maximum force that won't cause the block to slide up, then you want F to be smaller than W.
  • #1
dl447342
28
5
Homework Statement
You are sliding a block with mass m up a ramp inclined at an angle of theta with respect to the horizontal where the coefficient of static friction between the block and the ramp is mu_s. What is the maximum horizontal force you can apply to the ramp to prevent it from sliding up, assuming theta > 90 - theta_s, where theta_s is the angle between the ramp and horizontal when the force of static friction is maximum.
Relevant Equations
Wsin theta - F cos theta + f_s = 0,
-W cos theta - F sin theta + N = 0
I tried using Newton's first law as the net force in both the x and y directions should be zero in this case. In the free body diagram you need to consider weight, friction, normal force, and the horizontal force. I got a result that said that the horizontal force F you apply is at most W(mu_s + tan theta)/(1 - mu_s tan theta), where W is the weight. But this is negative for theta > 90 - theta_s. What does that say about the maximum horizontal force you can apply?
 
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  • #2
:welcome:

Is there a diagram you could post for this problem? I'm a bit confused: are you applying a force to the ramp or the block on the ramp? Also, I don't understand what ##\theta_s## is?
 
  • #3
PeroK said:
:welcome:

Is there a diagram you could post for this problem? I'm a bit confused: are you applying a force to the ramp or the block on the ramp? Also, I don't understand what ##\theta_s## is?
I can answer the second question but not the first. Angle θs is the angle of repose.
 
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  • #4
dl447342 said:
Homework Statement:: You are sliding a block with mass m up a ramp inclined at an angle of theta with respect to the horizontal where the coefficient of static friction between the block and the ramp is mu_s.
What is the maximum horizontal force you can apply to the ramp to prevent it from sliding up, assuming theta > 90 - theta_s, where theta_s is the angle between the ramp and horizontal when the force of static friction is maximum.
A. If you are sliding the block, why is static friction interesting? Should it say you are pushing horizontally on a block with mass m on a ramp?
B. You don't need to apply any force to stop it sliding up the ramp. Gravity will take care of that. So either we want the minimum force to prevent its sliding down, or the maximum force that won't result in its sliding up.
C. As you observe, if ##\theta+\theta_s>90°## then it will not slide up regardless of how hard you push. My guess is that it intends ##\theta+\theta_s<90°## but it could be a trick question.
 
  • #5
haruspex said:
A. If you are sliding the block, why is static friction interesting? Should it say you are pushing horizontally on a block with mass m on a ramp?
B. You don't need to apply any force to stop it sliding up the ramp. Gravity will take care of that. So either we want the minimum force to prevent its sliding down, or the maximum force that won't result in its sliding up.
C. As you observe, if ##\theta+\theta_s>90°## then it will not slide up regardless of how hard you push. My guess is that it intends ##\theta+\theta_s<90°## but it could be a trick question.
Oh yeah you're pushing horizontally on a block of mass m. My bad. I know that if $$ \theta + \theta_s > 90^\circ$$, the block won't slide up no matter how hard I push horizontally, but I just don't understand why that's the case. For example, can one show this mathematically? Is it related to the negative value of $$F^{max}$$ you get?
 
Last edited:
  • #6
dl447342 said:
Oh yeah you're pushing horizontally on a block of mass m. My bad. I know that if [tex] theta + theta_s > 90^circ [\tex], the block won't slide up no matter how hard I push horizontally, but I just don't understand why that's the case. For example, can one show this mathematically? Is it related to the negative value of [tex]F^max[\tex] you get?
So you are pushing on the block, not on the ramp, with a horizontal force. Let's see the free body diagram that you drew and the equation following from it that you wrote.
 
  • #7
kuruman said:
So you are pushing on the block, not on the ramp, with a horizontal force. Let's see the free body diagram that you drew and the equation following from it that you wrote.

image for applied force quesiton.png

Above is my FBD. I got the following equations:

x: $$f_s - F\cos \theta + W\sin \theta = 0$$
y: $$N - F\sin \theta -W\cos\theta = 0$$

Using the fact that $$f_s \leq \mu_s N$$ gave the result

##F \leq \frac{W(\mu_s + \tan \theta)}{1 - \mu_s \tan \theta}##
 
Last edited:
  • #8
How about the equation that follows from this?
 
  • #9
Nvm I get it. I end up with the result: ##F(1-\mu_s \tan \theta)\leq W(\tan\theta + \mu_s)## and so when I divide both sides by ##1-\mu_s \tan\theta##, I just swap the inequality sign so that any value of ##F## that's larger than ##\frac{W(\tan \theta + \mu_s)}{1-\mu_s\tan\theta}## will work.
 
  • #10
dl447342 said:
... I just swap the inequality sign ...
Why? Is ##1-\mu_s \tan\theta## always negative?
 
  • #11
kuruman said:
Why? Is ##1-\mu_s \tan\theta## always negative?
It's negative if ##\theta > 90^{\circ}-\theta_s##.
 
  • #12
dl447342 said:
It's negative if ##\theta > 90^{\circ}-\theta_s##.
I was asking for a proof of why that is always the case when ##\theta > 90^{\circ}-\theta_s##. It seems that you may not be clear about this considering the edited post #6 where you say that ##F \leq \frac{W(\mu_s + \tan \theta)}{1 - \mu_s \tan \theta}## and post #9
dl447342 said:
Nvm I get it. I end up with the result: ##F(1-\mu_s \tan \theta)\leq W(\tan\theta + \mu_s)## and so when I divide both sides by ##1-\mu_s \tan\theta##, I just swap the inequality sign so that any value of ##F## that's larger than ##\frac{W(\tan \theta + \mu_s)}{1-\mu_s\tan\theta}## will work.
where the inequality, although not written explicitly, is implied to be the other way.
 

Related to Maximum force applied to prevent block from sliding up ramp

1. What is the maximum force applied to prevent a block from sliding up a ramp?

The maximum force applied to prevent a block from sliding up a ramp is known as the static friction force. It is calculated by multiplying the coefficient of static friction (μs) by the normal force (N) acting on the block.

2. How is the coefficient of static friction determined?

The coefficient of static friction is determined by conducting experiments where the block is gradually tilted on the ramp until it starts to slide. The angle at which the block begins to slide is known as the angle of repose, and the coefficient of static friction can be calculated using the tangent of this angle.

3. What factors affect the maximum force needed to prevent a block from sliding up a ramp?

The maximum force needed to prevent a block from sliding up a ramp is affected by the coefficient of static friction, the weight of the block, and the angle of the ramp. A rougher surface, a heavier block, and a steeper ramp will require a higher maximum force.

4. Can the maximum force applied to prevent a block from sliding up a ramp be exceeded?

Yes, the maximum force of static friction can be exceeded if an external force, such as a push or pull, is applied to the block. This will cause the block to overcome the static friction force and start sliding up the ramp.

5. How does the maximum force needed to prevent a block from sliding up a ramp change if the ramp is inclined at different angles?

The maximum force needed to prevent a block from sliding up a ramp will increase as the angle of the ramp increases. This is because the steeper the ramp, the greater the component of the block's weight acting parallel to the ramp, which increases the force of static friction needed to prevent the block from sliding.

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