# 61 in Examkrackers Physics textbook

#### tinylilron

##### Member
10+ Year Member

Boards X and Y are massless and 4m in length. A 4N force F is applied to board Y as shown. Board X is held stationary. The two boards are nailed together at 1m from the left end of board Y. If the boards do not move, what is the static frictional force between the nail and board X?
A. 4N
B. 8N
C. 12N
D. 16N
For this problem, I did not know where to begin at first. I choose the nail as the point of rotation. So 3m*4N=1m*F, which would equal F=12N which is C.

However, the answer is D. EK says The net torque must be zero. If we choose the rotation point to be the end of the board Y, then 1m*F=4m*4N and F=16N, D.
EK also includes a little hint A little tip for torque problems: Find a force that is neither known nor asked for, and choose the point where this force acts to be your rotational point.

I guess my two major problems with this question. I was not really clear what it was asking me. How do I know that is the static frictional force? And I seem to have a huge issue with choosing the point of rotation.

Can anyone help me and shed some light on these crazy torque problems.

Oh and one random further question Tension=mass*force of gravity, right?

#### kzhan11

10+ Year Member
I initially stalled on this question until I realized that board X is being held stationary. So, picture the question this way: you have two things stuck together by a nail. Someone is trying to pull off the top piece by prying it off, but that fails. The nail MUST be exerting an opposite torque against the force you applied. However, the nail's torque has a different lever arm--if you set the point of rotation at the edge of Y, the lever arm for the nail is one meter and torque F's is four meters. Try the calculation that way, you should get 16N.

#### Got Em

15+ Year Member
Does anybody have a better explanation? I'm still lost after the first explanation.

#### lorenzomicron

I'm not sure I get the picture. are both boards horizontal, with X and Y overlapping? also where does the 4N force act? without this info, i could not possibly solve it.

#### surag

##### kobayashi
10+ Year Member
7+ Year Member
Is torque even on the mcat.....

#### Phantastic

10+ Year Member
The nail cannot be considered a point of rotation, just a point where a force acts. One force pushes up the board from the end. Imagine board Y as just being a door. When you push it open, you push farthest from the point of rotation to maximize the lever arm.

Now imagine board X is a plank that has been nailed over the door (it's like you're looking down at a zombie-proofed door). The door still wants to rotate at it's hinges, but now we have another force in addition to your attempt at pushing the door open: this force resists the push, because board Y isn't moving (whenever you see or think torque, think rotational equilibrium, torque(cw) = torque(ccw)). This opposing force acts 1m from the point of rotation (hinges).

Hope that helps.

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#### Phantastic

10+ Year Member
Is torque even on the mcat.....

Yes, and it is usually involved in situations of static equilibrium. Angular momentum is absent from the MCAT.

#### Got Em

15+ Year Member
I'm not sure I get the picture. are both boards horizontal, with X and Y overlapping? also where does the 4N force act? without this info, i could not possibly solve it.

Both Y and X are horizontal. Y is on top of X, but not exactly on top. Both boards are 4 m in length. The nail goes through both Y and X, where the LEFT END of Y is 1 m from the nail and the left end of X is about 3 m from the nail. The 4 N force is upwards and applied to the RIGHT end of Y.

Hope this is a little more clear. I'm trying to see if I can upload a picture.

#### Got Em

15+ Year Member
The nail cannot be considered a point of rotation, just a point where a force acts. One force pushes up the board from the end. Imagine board Y as just being a door. When you push it open, you push farthest from the point of rotation to maximize the lever arm.

Now imagine board X is a plank that has been nailed over the door (it's like you're looking down at a zombie-proofed door). The door still wants to rotate at it's hinges, but now we have another force in addition to your attempt at pushing the door open: this force resists the push, because board Y isn't moving (whenever you see or think torque, think rotational equilibrium, torque(cw) = torque(ccw)). This opposing force acts 1m from the point of rotation (hinges).

Hope that helps.

Sorry, did not see your attachment.

Now I understand this a little better, but I still have a few questions. How come the distance is not used from the left end of X? Also, in the explanations it states that "Tip: Find a force that is neither known nor asked for, and choose the point where this force acts to be your rotational basis.". How do I find this force is it is not known? I mean they have a force of 12N upwards, but how was I supposed to know this when doing the problem? Thanks in advance.

#### lorenzomicron

Ok, so the idea behind this is that the whole structure doesnt move. Thus, the force F can be moved (translated) from the end of the board to the point of the hinge. This is a point 4 meters away from the end of the x-board. now, the F force causes a reactionary force in the y board of 4 N down. but, this force is not at the hinge but at the end of the x-board. at the hinge, a torque is causes on the x board. this torque is 16 N in magnitude (4x4), which is opposed by a torque which need to act 1 m from the end of the Y board. Thus, 16x1 = 16. I think this make sense.

#### lorenzomicron

actually, the above answer left some important things out.

the way to deal with multi-structure problems is to analyze each part separately in static equilibrium. so, for plank Y, we get two forces, -4 N at one end, and 4 N at the other end. now, it is in static equilibrium. if we take the torque at the end of plank Y, we then get 16 Nm.

let's say that plank X must then supply the force that counteracts this 16 Nm torque. so, in order to counteract this torque at the 1 m point from the end of plank Y, what force do we need? we need 16Nm/1m= 16 N force.

#### fizzgig

##### LudicrousSpeed!
7+ Year Member
people have covered it above but for the sake of hearing it 10 different ways (which helps me sometimes)...

only look at board X - it's the board in question. draw a free body diagram for board X (meaning only draw board X and draw ANY forces acting on it... any other object you remove will leave its force on board X behind on the free body diagram. it's nailed to a board that's not allowed to move.

so you have a force on the right end (F=4N), a weight (yay, massless, we get to leave this one out, the nail (let's come back to this, we'll assume we dont know anything about it yet), and the force that board Y exerts on X (don't know this either).

by mentally prying board X up a bit, you can see that if board X is rigid and Y does not give at all, X will have to rotate about its leftmost point, pulling out the nail, if X is going to move. now you know the location of the force board Y will exert. its magnitude you don't know, but it logically will be pointing up.

now the nail: they ask for the static friction force, so you know nothing in the system is moving bc of these applied forces. you also know that the force here is acting between the nail surface and the wood surface - so this force must either point up or down.

now you know locations for 3 forces and magnitude for 1. since the board is not moving, you can place your solving point ANYWHERE and the moments (force*distances, tendency to rotate about your solving point) MUST sum to zero, or suddenly the board is not really in static equilibrium and it will rotate.

for convenience, we choose the point on the far left - we don't know this force and we don't want to solve for it. now that force goes away because the distance to it is zero.

you're left with 1m*NailForce = 4m*AppliedForce

NailForce = 16N

if you chose nailforce to be in the wrong direction, no biggie, it'll be negative, so you don't even have to spend loads of time making sure you 'get' which direction to make that force point.

#### IlyaR

5+ Year Member
since the board is not moving, you can place your solving point ANYWHERE and the moments (force*distances, tendency to rotate about your solving point) MUST sum to zero, or suddenly the board is not really in static equilibrium and it will rotate.

This is where I messed up. I placed the solving point directly in the middle of board Y and ended up getting 8N as the static frictional force.

What am I doing incorrectly?

If the point of rotation is in the middle of Y, then the force on the right side of Y generates
2M (4N)= 8NM= 1M (Nail Force)= 8N

Intuitively I understand that if you pry the right side up, the hinge will be on the left side, thus 4M(4N)= 1M( Nail Force), but from my understanding you can choose the point of rotation anywhere and still arrive at the correct answer, is this wrong?

Thanks!

#### sniderwes

7+ Year Member
I was also really confused about this problem. The way I went about it was by examining upward and downward forces on board Y.

The problem says the boards do not move, so upward forces have to equal the downward forces. The only upward force is F, 4N. The only downward force is static friction from the nail, so wouldn't it have to equal 4N?

#### inasensegone

This is a torque problem because you are trying to prevent rotation (lifting up the right side of the board where the force of 4 N is applied). So you must use Force1 * lever arm1 = F2 * lever arm 2.

Make the applied 4 newton force F1. This force is upwards (which again would cause the board to rotate counter clockwise with the far left side of the board acting as the point of rotation). Because the far left side of the board is the point of rotation, the lever arm for the 4 newton force is 4 meters. Now we have one side of our equation: 4N * 4 meters = F2LA2

For the other side of our equation we need to generate an equal (and opposite) amount of torque. This torque is provided by the nail. We know the nail is 1 meter from the left side of the board (the point of rotation) so the equation becomes: 4N * 4 M = F2 * 1 M. Solve for F2 and you get 16 newtons

Edit: Understand that these two torques are equal to each other because they must cancel out to prevent rotation (lifting of one side of the board).

Picture the classic door example: If you push on a door where the doorknob is, you can push open a door with a normal amount of force. If you apply that same amount of force closer to the point of rotation, that is, closer to the hinges, then it will be more difficult to open the door. You haven't changed the amount of force you're applying, you only changed where you are applying it. In a nutshell, this is why the answer can't also be 4 Newtons.... the forces are applied at different points on the board. (Try the door example at home if you need to prove this to yourself because it helps with good conceptualization)

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#### sniderwes

7+ Year Member
This is a torque problem because you are trying to prevent rotation (lifting up the right side of the board where the force of 4 N is applied). So you must use Force1 * lever arm1 = F2 * lever arm 2.

Make the applied 4 newton force F1. This force is upwards (which again would cause the board to rotate counter clockwise with the far left side of the board acting as the point of rotation). Because the far left side of the board is the point of rotation, the lever arm for the 4 newton force is 4 meters. Now we have one side of our equation: 4N * 4 meters = F2LA2

For the other side of our equation we need to generate an equal (and opposite) amount of torque. This torque is provided by the nail. We know the nail is 1 meter from the left side of the board (the point of rotation) so the equation becomes: 4N * 4 M = F2 * 1 M. Solve for F2 and you get 16 newtons

Edit: Understand that these two torques are equal to each other because they must cancel out to prevent rotation (lifting of one side of the board).

Picture the classic door example: If you push on a door where the doorknob is, you can push open a door with a normal amount of force. If you apply that same amount of force closer to the point of rotation, that is, closer to the hinges, then it will be more difficult to open the door. You haven't changed the amount of force you're applying, you only changed where you are applying it. In a nutshell, this is why the answer can't also be 4 Newtons.... the forces are applied at different points on the board. (Try the door example at home if you need to prove this to yourself because it helps with good conceptualization)
Thanks, your way does make sense thanking about it as a door =]
The only thing I'm still confused about is where the 12 N (at the "hinge") comes from. I know you can ignore it for torque if you set your point of rotation there, but what is this upward force from? Is it just board Y pushing down on board X when rotating, and board X pushing back on it?

#### inasensegone

Thanks, your way does make sense thanking about it as a door =]
The only thing I'm still confused about is where the 12 N (at the "hinge") comes from. I know you can ignore it for torque if you set your point of rotation there, but what is this upward force from? Is it just board Y pushing down on board X when rotating, and board X pushing back on it?

No problem.

I'm not sure I completely understand what you mean? I think what you are asking, is where is the 12 Newtons coming from, given that you only supplied 4 Newtons upward (counterclockwise) on the right side initially? Well it isn't an extra 12 newtons... remember, you are supplying 4 newtons and this is supplied at 4 meters distance from the point of rotation. Therefore, you are supplying 16 N* M of torque. That is, Newton-meters. So there isn't an extra 12 newtons.

The nail on the other hand is much closer to the point of rotation so it's component of lever arm (in meters) is much smaller than the lever arm from the person lifting the board up on the right side. This is why the force that needs to be supplied by the nail is so much greater. Again, try to relate it to the door example.

Picture 2 people on opposite sides of a door trying to push the door toward each other. If one dude is pushing near the doorknob, in order to cancel him out, the other person can apply the same force on the opposite side of the door at the doorknob, or a much greater force will be needed as he pushes more closely to the hinges. This has to do with the effective lever arm becoming shorter.

Let me know if that makes sense.

#### sniderwes

7+ Year Member
No problem.

I'm not sure I completely understand what you mean? I think what you are asking, is where is the 12 Newtons coming from, given that you only supplied 4 Newtons upward (counterclockwise) on the right side initially? Well it isn't an extra 12 newtons... remember, you are supplying 4 newtons and this is supplied at 4 meters distance from the point of rotation. Therefore, you are supplying 16 N* M of torque. That is, Newton-meters. So there isn't an extra 12 newtons.

The nail on the other hand is much closer to the point of rotation so it's component of lever arm (in meters) is much smaller than the lever arm from the person lifting the board up on the right side. This is why the force that needs to be supplied by the nail is so much greater. Again, try to relate it to the door example.

Picture 2 people on opposite sides of a door trying to push the door toward each other. If one dude is pushing near the doorknob, in order to cancel him out, the other person can apply the same force on the opposite side of the door at the doorknob, or a much greater force will be needed as he pushes more closely to the hinges. This has to do with the effective lever arm becoming shorter.

Let me know if that makes sense.
Ah I'm sorry, I didn't make it clear what 12 N I was referencing. I've attached EK's answer diagram below, as you can see they have a 12 N force coming from the "hinge" of our door. My reasoning was that unlike the hinge of a regular door, the hinge in our example is merely the point where two boards meet. So when our door tries to swing open, the board underneath it will push back on it upwards, generating the 12 N. But this doesn't apply to normal hinges, because regular doors freely rotate.

I'm not sure if that's correct though, and it's the only explanation I could come up with for why there would be the 12 N of upward force...

#### inasensegone

Hrmmm.. that's a good question. I imagine that this was EK's convoluted way of explaining 2 approaches.

If you take the nail to be the point of rotation instead, 4 N * 3 m = 12N * 1 m

The math still works, but I'm not sure what else EK might have been trying to illustrate here.

They may have also been trying to illustrate that if the 16 newtons is supplied by the nail (downward), then a normal force of 12 newtons (upward) would be provided by the stationary board underneath to counteract this 16 newtons, given that 4 newtons was already supplied from the underside.

I wouldn't worry too much about it, as long as you understand a quick approach to similar problems and understand the concepts and the bigger picture. Sometimes Berkeley review's explanation doesn't make sense to me, but I will find a way that works for me and is fast and intuitive, and that's all that matters for the MCAT.

#### sniderwes

7+ Year Member
Yeah...sometimes their explanations can be a bit confusing haha. I just did their first sectional practice test and got a projected score of 10, admittedly with some careless mistakes. Sigh, just gotta get through this year however ****ty it might seem.

Thanks for working with me on this problem! =]