Pushing down on a block at an angle on a horizontal table with friction

[RekaNice]

Homework Statement

Example 2.12 : A block pulled on a horizontal table with friction. A block of weight 100 Newtons rests on a horizontal table. The coefficient of static friction between the block and the table is µs = 0.8.
a. If someone pulls on the block with a force F in the direction 37◦ above the horizontal, what is the minimum value of F which will cause the block to slip?
b. If someone pushes on the block with a force F in a direction 37◦ below the horizontal, what is the minimum value of F which will cause the block to slip?
c. Same question as (b), except the direction of the push is 53◦ below the horizontal.

Homework Equations

Carrying out a similar analysis for part (c), we obtain f = F cos 53◦ = .6F and N = (100 Newtons) + F sin 53◦ = 100 + .8F and thus f N = .6F 100 + .8F (2.7) If we again draw a graph of f /N as a function of F (Fig. 2.21) we see that the limiting value of f /N, as F → ∞, is 0.75. Thus it is clear that the ratio f /N will never attain the value 0.8 no matter how large F is. Therefore it is impossible to cause the block to slip by pushing on it in a direction 53◦ below the horizontal. As we push harder and harder, the normal force magnitude N increases fast enough so that the plane can always exert a large enough frictional force to counterbalance the horizontal component of the applied force F~ . If we set f /N (as given by eqn.( 2.7)) equal to 0.8 and solve for F, we obtain F = −2000 Newtons. Does this negative value of F have any physical significance? The obvious guess is that a negative push should be interpreted as a pull and that we have shown that a pull of 2000 Newtons in a direction 53◦ above the horizontal will just suffice to make the block slip. This is incorrect! If we repeat the analysis of part (a) for the case when the applied force F is a pull in the direction 53◦ above the horizontal, we find that the critical value for slipping is F = 64.5 Newtons. We conclude that the value F = −2000 Newtons has no physical significance. This illustrates the usefulness of drawing graphs like Fig. 2.21 rather than just formally solving equations. Mathematically, the reason why a negative value of F does not correspond to a pull is that when we replace the symbol F by −F the equations describing case (a) do not go over into those describing case (b).

The Attempt at a Solution

I understand how to solve the problem but what I don't understand here is why a negative pushing force here can't be interpreted as a pull?

 

[Let'sthink]

Friction is a self adjusting force both in direction and magnitude and its direction is always opposite to the motion or tending motion.

 

[haruspex]

The correct equation for static friction is $|F_{fric}|≤\mu N$, where N ≥0.
In case a), with F defined so as to be positive, this yields $|F_{fric}|=\mu|mg-F\sin(\theta)|=|F\cos(\theta)|=F\cos(\theta)$.
In b), with F again defined so as to be positive, it gives $|F_{fric}|=\mu|mg+F\sin(\theta)|=F\cos(\theta)$.
If we try switching the sign of F in case a) so as to treat b) as a negative pull we get:
$|F_{fric}|=\mu|mg+F\sin(\theta)|=-F\cos(\theta)$, so not the same equation.

Interestingly, the text is wrong to say that switching the sign has no physical significance. You can see that switching the sign of g as well compensates. So the 2000N is the solution for the block being pushed up at an angle of 53 degrees above the horizontal against the underside of the table.

 

[Let'sthink]

Positive and negative are conventional terms. You decide which direction is positive. The opposite direction to that is negative and then you live with it. Whereas Push and Pull are subjective term it depends on the subject. For the subject pushing direction may be considered as positive according to his convention then pull will be negative. But this is a local convention. This needs to be translated into universal conventions of +x, -x and +y and -y, because all other forces are written or can be written out in that convention.

 

[David Roberts]

Push and pull are two different types of forces. As most engineers know, you can't push on a rope or at least you shouldn't expect to get anything moved. Not disagreeing, just adding a nuance.

 

[haruspex]

Push and pull are two different types of forces. As most engineers know, you can't push on a rope or at least you shouldn't expect to get anything moved. Not disagreeing, just adding a nuance.

Ok, but that does not have any relevance to the question posed here.

 

[Let'sthink]

Push and pull are two different types of forces. As most engineers know, you can't push on a rope or at least you shouldn't expect to get anything moved. Not disagreeing, just adding a nuance.

I agree, you cannot push a rope. Although push and pull refers to the elementary oldest idea to define force. These terms are situational and generalizing that push and pull are different forces may not be in order. I would say push gives rise to compression stress and pull gives rise to tensile stress. Not only they give rise to but they generate the push and pull as restoring phenomenon.

 

[RekaNice]

The correct equation for static friction is $|F_{fric}|≤\mu N$, where N ≥0.
In case a), with F defined so as to be positive, this yields $|F_{fric}|=\mu|mg-F\sin(\theta)|=|F\cos(\theta)|=F\cos(\theta)$.
In b), with F again defined so as to be positive, it gives $|F_{fric}|=\mu|mg+F\sin(\theta)|=F\cos(\theta)$.
If we try switching the sign of F in case a) so as to treat b) as a negative pull we get:
$|F_{fric}|=\mu|mg+F\sin(\theta)|=-F\cos(\theta)$, so not the same equation.

Interestingly, the text is wrong to say that switching the sign has no physical significance. You can see that switching the sign of g as well compensates. So the 2000N is the solution for the block being pushed up at an angle of 53 degrees above the horizontal against the underside of the table.

Fyi, this text is from a free online textbook: Classical Mechanics-A Critical Introduction. So, what u r saying is the force is pushing from below? It makes sense but is rather confusing at the same time, are u really sure we can do that? And is there a reason why I can't interpret it as a pull?

I also noticed that F(push) will be negative when the angle gets bigger. Why is this so conceptually? Is it bcos the Normal force will be too great then to the point that pushing downwards at an angle is futile due to the bigger maximum static friction, and that instead pushing upwards at an angle is better to make the block slip