Maximum force applied to prevent block from sliding up ramp

[dl447342]

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?

 

[PeroK]

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?

 

[kuruman]

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 θ_s is the angle of repose.

 

[haruspex]

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.

 

[dl447342]

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?

 

[kuruman]

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 $$F^max[\tex] you get?$$[/tex]

[tex]$$So you are pushing on the block, not on the ramp, with a horizontal force. Let's see the free body diagram that you drew and the equation following from it that you wrote.$$[/tex]

 

[dl447342]

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.

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

$F \leq \frac{W(\mu_s + \tan \theta)}{1 - \mu_s \tan \theta}$

 

[kuruman]

How about the equation that follows from this?

 

[dl447342]

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.

 

[kuruman]

... I just swap the inequality sign ...

Why? Is $1-\mu_s \tan\theta$ always negative?

 

[dl447342]

Why? Is $1-\mu_s \tan\theta$ always negative?

It's negative if $\theta > 90^{\circ}-\theta_s$.

 

[kuruman]

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

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.