# Friction forces to keep board from slipping

1. Jun 10, 2006

### rcwha

The board sandwiched between two other boards weighs 95.5 N. If the coefficient of friction between the boards is .663, what must be the magnitude of the comrssion forces (assume horizontal acting on both sides of the center board to keep it from slipping?

I have no idea how to do this problem...is the normal force 95.5 (if so then i can solve for friction but i wouldnt know what to do then...The answer is 72 N..so if somebody can show me how to do this problem i would greatly appreciate is.....
thank you

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2. Jun 10, 2006

### Hootenanny

Staff Emeritus
I can't see your attachment yet, but I'm assuming that this sandwhich of boards lies in the vertical plain, perpendicular to the horizontal? Or is it lying at an angle?

3. Jun 10, 2006

### rcwha

the sandwhich of boards lies in the vertical plain...i added an http://

4. Jun 10, 2006

### Hootenanny

Staff Emeritus

Last edited by a moderator: May 2, 2017
5. Jun 10, 2006

### Saketh

Picture it this way - it is a battle between friction and gravity. As you already said, gravity is pulling down with a force of 95.5N. In order for the center plane to stay up, friction has to push up with an equal force. Since there are two planes, there are two different forces of friction that add up to be 95.5N.

Use your knowledge of friction forces and the fact that the friction forces have to add up to be the weight of the center plane to find the compression force.

I'm assuming you mean "compression forces," not "comrssion forces".

6. Jun 10, 2006

### rcwha

so i dont have to wrry about the horizontal forces? i thought the compressive forces meant the horizontal forces....anyways this is what i did...i found friction to = 60.45 by multiplying 95.5,which im asumming is the normal force time .633 (coef. friction). So the board is slipping down with a force of 95.5 (normal force) - 65.27 (friction)= 30.23

so what do i do from here, and how do i relate Fy to Fx... I understand fy and Fx has to = 0 b/c the acceleration is 0 but what now?

7. Jun 10, 2006

### Saketh

The compressive forces act in the horizontal direction - you are correct. But the frictional forces act in the vertical direction.

The weight of the center plane is not the normal force for friction. Let us say that $$F_{friction}=\mu F_{N}$$. The weight of the object is not $$F_{N}$$.

Think about what is the normal force. It's not the weight of the object. The normal force is always perpendicular to the frictional force, so think about what force in this problem is perpendicular to the frictional force. If friction is acting vertically, then what is acting horizontally? Once you figure this out, use $$\Sigma F_{y} = 2F_{FR}-95.5$$ as your net-force equation in the y-direction. Since the plate is sandwiched between two other plates, there are two forces of friction. You also know that $$F_{FR}=\mu F_{N}$$. All you need now is $$F_{N}$$.

Let us say that there is no compressive force. The center plane would fall, regardless of its weight. Let us say that there was a gigantic compressive force. What would happen? What can you conclude about the normal force?

8. Jun 11, 2006

### rcwha

...that the normal force is zero?

9. Jun 11, 2006

### Hootenanny

Staff Emeritus
Let us get back to the question in hand. You have two surfaces which can create friction, lets call the left surface of the middle block A, and the right surface B. Now, you have a downward force of 95.5N; to stop the middle board from falling the total upward force must equal 95.5N. This upward force is supplied by the two frictions;

$$F_{A} + F_{B} = 95.5$$

Now because the force applied to each outerboard is equal, we can say that $F_{A} = F_{B} = F \Rightarrow F_{A} + F_{B} = 2F$, thus;

$$2F = 95.5$$

Where F is the frictional force. Do you follow?

10. Jun 11, 2006

### rcwha

yeah it makes sense now...thank you guys

11. Jun 11, 2006

### Hootenanny

Staff Emeritus
My pleasure