Pushing a book against a wall and keeping it up

[Stewie117]

Homework Statement

A book of mass M is positioned against a vertical wall. The coefficient
of friction between the book and the wall is μ. You wish to keep
the book from falling by pushing on it with a force F applied at an
angle θ with respect to the horizontal (−π/2 < θ < π/2).
For a given θ, what is the minimum F required?

Relevant Equations

Sum of horizontal forces =0, sum of vertical forces =0.

The answers show that the static friction between the book and the wall points in the same direction as the vertical component of the applied force. That is, Fsin(θ) + ForceFriction -Mg=0. But why does the friction point in the same direction as the vertical component of F? More generally, if there is more than 1 force acting on an object (as in this case), how can I find the direction of the friction?

 

[Halc]

The answers show that the static friction between the book and the wall points in the same direction as the vertical component of the applied force. That is, Fsin(θ) + ForceFriction -Mg=0.

Yes. Two up (positive, being added), one down (being subtracted.

But why does the friction point in the same direction as the vertical component of F?

Because the wall friction is exerting an upward force on the book, just like the vertical component of F is also exerting an upward force on the book. Only gravity is exerting a downward force on the book, and the two upward forces are just enough to cancel that.

If F is high enough, it will be more than Mg and the friction acting on the book will become negative, but you're asked to find the minimum force, not the maximum.

 

[Lnewqban]

... But why does the friction point in the same direction as the vertical component of F? More generally, if there is more than 1 force acting on an object (as in this case), how can I find the direction of the friction?

We call "friction" to that force (exerted by each surface on the other) that resists the relative movement between two surfaces in contact.
The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

Please, see:
https://en.wikipedia.org/wiki/Friction

 

[Stewie117]

We call "friction" to that force (exerted by each surface on the other) that resists the relative movement between two surfaces in contact.
The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

Please, see:
https://en.wikipedia.org/wiki/Friction

Yes. I understand what friction is conceptually. But, in this you can't determine the direction of the net applied force since there are two forces two forces acting vertically on the book, Fsin(θ) and Mg and the magnitude of F is unknown. I get that if only Mg were acting on the book, the friction would oppose the downwards movement of the book. But, in this problem, we have Fsin(θ) too. I'm not sure why ForceFriction would point in the same direction as Fsin(θ). The friction could oppose Fsin(θ) or Mg, and I'm not sure which of those it has to oppose. You get what I'm saying?

 

[Lnewqban]

Yes. I understand what friction is conceptually.
...You get what I'm saying?

Sorry, I could not know how much you knew.
I don't fully get it, but will keep trying.
In real life, you only are dealing with two forces acting on the book: the one from gravity (vertical direction) and the one from the hand (diagonal direction).

Let's find out together how extreme changes in the direction of the application of the hand force (same magnitude) affect friction:
1) Pure vertical direction.
2) Pure horizontal direction (pushing the book against the wall).

 

[etotheipi]

Yes. I understand what friction is conceptually. But, in this you can't determine the direction of the net applied force since there are two forces two forces acting vertically on the book, Fsin(θ) and Mg and the magnitude of F is unknown.

I think I understand your concern. If we don't know the relative sizes of the other two forces, generally we can't say the direction of friction.

However, as @Halc and @Lnewqban have pointed out in this case, you're after the minimum force $F$ here. This implies that if $F$ were just a tiny bit (infinitesimally) smaller for a given $\theta$, the book would start slipping down (assuming $\theta < \frac{\pi}{2}$ as shown in the diagram). The friction is going to point upward to oppose this slippage.

 

[kuruman]

The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

Nope. Static friction is whatever is necessary to provide the observed acceleration, in this case zero. Imagine varying the magnitude of $F$ at will, but still keeping the book at rest against the wall. If the vertical component of the pushing force $F$ is equal to the weight, no static friction is needed to keep the book in place. If the vertical component of $F$ is greater than the weight, then static friction will point down so that the net vertical force is zero. Likewise, if the vertical component of $F$ is less than the weight, then static friction will point up so that the net vertical force is still zero. Here we are looking for the minimum value of $F$ which means that static friction must point up and have its maximum value. If $F$ is reduced below that minimum value, the book will slide down.

 

[etotheipi]

The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

If the vertical component of the pushing force $F$ is equal to the weight, no static friction is needed to keep the book in place. If the vertical component of $F$ is greater than the weight, then static friction will point down so that the net vertical force is zero. Likewise, if the vertical component of $F$ is less than the weight, then static friction will point up so that the net vertical force is still zero.

I think you are perhaps in agreement, there is just a difference in terminology. I believe when @Lnewqban refers to the 'net applied force' what is actually meant is the total force in the direction of the interface of all forces minus the static friction. So in this case, the 'net applied force' is referring to that resultant force of magnitude $F\sin{\theta} - mg$.

 

[kuruman]

I think you are perhaps in agreement, there is just a difference in terminology. I believe when @Lnewqban refers to the 'net applied force' what is meant is the total force in the direction of the interface of all forces (before friction is added in). So in this case, the 'net applied force' is referring to that resultant force of magnitude $F\sin{\theta} - mg$.

I beg to differ. If you draw the FBD of the book, there are three forces acting on it, (1) the weight, (2) the contact force at the wall and (3) force $F$ - that's it. Now you can add all three of them vectorially and get the "net" force or you can give names to components like "friction", "normal", "vertical", "horizontal", etc. However, it is a stretch to cherry pick the sum of the vertical components of two of the forces but not the third and call it the net applied force. Also note the wording in the statement of the problem, "$\dots$ by pushing on it with a force F applied at $\dots$" It stands to reason to call $F$ and nothing else the applied force.

 

[etotheipi]

You're right, I just interpreted it slightly differently. I have found it a very useful problem solving technique sometimes to imagine removing the static friction force and determining what motion would result, in order to ascertain the direction of the frictional force.

Such an approach is useful for e.g. analysing rolling motion. For a cylinder rolling down a hill under a gravitational force, without the frictional force the linear velocity would increase whilst the angular velocity would remain constant. Hence we insert the frictional force backward to complete our free body diagram.

I would agree that 'net applied force' is a confusing terminology; I would call it explicitly 'net force minus static frictional force'. But the notion of imagining the motion without the static friction force and 'inserting' it subsequently is useful for problem solving.

 

[haruspex]

The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

Nope

It is worth rewording to make it exact, but I think it is clear @Lnewqban meant the net of all the other forces. Is that your only objection?
The rewording does highlight the question of what if there is more than one frictional force with a component in some direction? These cases are often statically indeterminate.

 

[kuruman]

It is worth rewording to make it exact, but I think it is clear @Lnewqban meant the net of all the other forces. Is that your only objection?
The rewording does highlight the question of what if there is more than one frictional force with a component in some direction? These cases are often statically indeterminate.

When the net force is zero, anyone force is always opposite to the net of all the other forces. That in itself is insufficient to determine the direction of the force of static friction. In order to specify the direction of friction, one needs the additional piece of information that the applied force $F$ is at a minimum. Whatever @Lnewqban may have meant (we need to hear from him on this), the statement

The direction of that force vector is always parallel to the surfaces and has a direction opposite to the net applied force.

other than the wording "applied force", i.e. $F$, instead of "sum of remaining forces" does not address OP's question,

But, in this problem, we have Fsin(θ) too. I'm not sure why ForceFriction would point in the same direction as Fsin(θ). The friction could oppose Fsin(θ) or Mg, and I'm not sure which of those it has to oppose. You get what I'm saying?

and clarify why the force of static friction has to point up in this case.

 

[haruspex]

When the net force is zero, anyone force is always opposite to the net of all the other forces. That in itself is insufficient to determine the direction of the force of static friction.

I took @Lnewqban's statement as generic for both static and kinetic.

In order to specify the direction of friction, one needs the additional piece of information that the applied force F is at a minimum.

That doesn’t make the statement false, just insufficient by itself in the present context.

 

[kuruman]

I took @Lnewqban's statement as generic for both static and kinetic.

I took that statement to apply to this problem where there is static friction at the threshold of sliding.

That doesn’t make the statement false, just insufficient by itself in the present context.

I agree. I never said that the statement is false. If anything, it is misleading in the sense that "is always parallel to the surfaces and has a direction opposite to the net applied force" makes one think that the direction can be found by figuring out the direction of the "net applied force", however one understands that. Anyway, I do not wish to belabor the point. We haven't heard from OP since post #4. It's OP's question and I believe we should wait to hear from OP whether the question has been answered to OP's satisfaction..

 

[etotheipi]

I took @Lnewqban's statement as generic for both static and kinetic.

Inserting the kinetic friction in the opposite direction to the net of other forces only works if the net of all other forces are acting to increase the relative velocity of the two surfaces undergoing slippage.

For example, we might consider a block sliding along a rough surface, being pulled in the negative direction by a rope in tension (slowing it down). The net of other forces (the tension) is in the negative direction, but the velocity of the ground w.r.t. the box is also in the negative direction and so is the friction. The net of other forces is acting to reduce the slippage, and the friction is in the same direction as this.

Even for static friction, a box might rest on a rough table of equal mass to the box, and we might apply equal horizontal forces to both the table and the box (the table is on a smooth surface) so that they share the same acceleration. The net other applied force on the box is non-zero, but the friction is zero since there is no relative motion.

 

[haruspex]

Inserting the kinetic friction in the opposite direction to the net of other forces only works if the net of all other forces are acting to increase the relative velocity of the two surfaces undergoing slippage.

Good point.

 

[Lnewqban]

I appreciate everyone's corrections.
Sorry, I did not mean to create additional confusion to the OP.
I just wanted him to clearly see the clamping effect of the applied force.

It seems a simple problem of vectors to me, which the OP may not be seeing clearly.
Please, consider that English is not my native language and that I am not a Physicist by any means.

At best, I have been doing some mechanical engineering work, and my more intimate relationship with friction has been intuitively sensing and estimating it in order to keep the tires of my motorcycles from skidding much from under me for many years of happy riding.

Again, suggestions and corrections from experts are always welcome, especially when those help better understanding of OP's in need of help.

 

[Lnewqban]

Just in case the OP comes back, or others use this thread as future reference, I am posting this link to a good article on the subject:

https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/