# What is the theory behind the following problems involving friction

1. Oct 26, 2013

### hbk69

Hi i have been trying to find a pattern between friction problems in-order to help me answer different scenarios correctly, i hope you can help my understanding please.

1) A box that weighs 10.0 N is being dragged with constant velocity along a horizontal surface of the table by a rope that is at an angle of 45 deg with that surface. The tension in the rope is 5.0 N. What is the coefficient of friction?

Ff= Friction force
Fg= weight

Fcosθ= component acting along the ramp
Fsinθ= component acting perpendicular to the ramp (Normal Force)

To calculate the Friction force we, Fg - Fsinθ

In this case why do we substract Fsinθ from Fg to work out the Ff?

2) A 20kg box sits on an incline of 300. If the coefficient of sliding friction between the box and incline is 0.3 what is the acceleration of the box down the incline?

Here,

mgsinθ= weight acting along the ramp
Fn the normal force = mgcosθ

I found the formula which makes it simple to solve the problem,

where Facceleration = mgsinθ - μkmgcosθ, with μk being the coefficient of kinetic friction.

QUESTION: In this problem why is the normal force mg "cosθ" because in question 1) the normal force was Fsinθ, also why is the μk coefficient of kinetic friction associated with mgcosθ when i need to find the Facceleration.

Also in this scenario can the Friction force be calculated the same way as question 1) by Fg - Fsinθ ? or is there a similar concept or way of finding it?

3) A 1200 kg car is coasting down a 30 degrees hill. When the car’s speed is 12 m/s the driver applies the brakes. What constant force F (parallel to the road) is required if the car is to stop after travelling 100m?

In this scenario can we use the Facceleration equation from question 2) to solve the problem (Facceleration = mgsinθ - μkmgcosθ) or a=gsinθ? if not what is the reason for this?

Overall Question

Are all these above question very similar in the sense that they share a fundamental concept which can help me solve them? or any other friction related problems involving inclined ramps etc and horizontal surfaces. And is there a simple method to help solve problems when looking at inclined problems and non inclided problems.

Thank you very much for any help guys i have been stressed out trying my best to understand friction related problems.

2. Oct 26, 2013

### Staff: Mentor

Here are the keys to doing a problem like this.

Step 1: Draw a free body diagram of the object moving down the incline (or moving horizontally in the case of no incline). This is the most important step.

Step 2: Identify all the forces that are acting on this free body, and draw vectors representing these forces on your free body diagram.

Step 3: Using geometry, resolve these vector forces into components perpendicular to- and parallel to the incline.

Step 4: Write down two equations for the components of the net force on the body in the directions perpendicular to and parallel to the incline.

Step 5: Recognize that there is no acceleration of your body in the direction perpendicular to the incline. Therefore, the net force in that direction must be equal to zero. Use this equation to determine the perpendicular component of force exerted by the incline on the body. This is called the "normal force."

Step 6: Set the component of the net force on the body in the direction parallel to the incline equal to ma, since the body may be accelerating down the incline. One of the terms in this equation will be the tangential force exerted by the incline on the body. This tangential force is called the frictional force. You don't know the value of the frictional force yet but you can determine it by multiplying the normal force from Step 5 by the coefficient of kinetic friction.

You now have enough information to solve for whatever you are asked to determine in the particular problem you are doing.

3. Nov 3, 2013

### tiny-tim

hi hbk69!
because the angle θ is a different angle in the two questions

in 1) θ is the angle between the surface and the rope …

so 90° - θ is the angle between the normal and the tension force (the rope)

in 2) θ is the angle between the surface and the horizontal

so θ is also the angle between the the normal and the weight force (the vertical)
no, Fg - Fsinθ was not the friction force, it was the normal force (N, which we divide by the friction force to get the coefficient of friction, µ)

this is because the only three forces acting on it are Fg friction and N

and the normal acceleration is zero

(this is very important … a lot of problems depend on the normal acceleration being zero)

so Fg - Fsinθ - N = 0

that is why Fg - Fsinθ = N

the friction force in 1) was calculated from the horizontal F = ma equation …

i'm not sure you've grasped this, so can you show us what you think that equation is?
almost everything in mechanics is writing out the appropriate F = ma equations

in the case of slopes, the fundamental thing to remember is that the normal acceleration is always zero (because the normal distance is always constant, ie 0), so you can write ∑ F = 0 (along the normal), and that helps you find N

once you've found N, µN give you the friction force, and you can then use F = ma along the slope

alternatively, if you're told that the acceleration along the slope is 0, then you know ∑ F = 0 (along the slope), and that gives you the friction

(you must stop trying to remember "friction formulas", and start using F = ma from scratch)​

4. Nov 4, 2013

### hbk69

So the Fg-Fsinθ comes from the F=ma horizonta but if so howl?

But going by what you have said then the equation would be Ff(friction force)=μK(kinetic coefficient)*N(Fg-Fsinθ), where Fsinθ is the vertical component and Fg the weight.

When you say the normal acceleration=0 what do you mean? as the normal is always vertical then the acceleration acting downwards is always constant at 9.8 so you are say the upwards acceleration is 0? sorry am stupid lol

When it comes to the problems involving a slope i think i am still lost. When i use F=ma along the slop and when N,μN are involved how would i include the angles,? i know that the Normal is upwards the weight is downwards and the friction force is in the opposite direction of the objects motion but how would i relate all of these to find out what the question asks me from the F=ma. I give you an example, i was reading a book on inclined planes and i found a number of equations but did not understand them properly:

Fpull-Fpush= ma

Fpush=mgsinθ+Ff(friction), where Ff=μsFn, where Fn=mgcosθ

Fpull=mgsinθ+usmgcosθ

Facceleration=mgsinθ - μkmgcosθ

What is the Fpull/Fpush and the associated variables such as mgsinθ, cos etc must be the different forces acting on the relevant object. I want to be able to read the situation correctly and automatically know how to derive the correct formula from F=ma when it come to the slopes

Also could you show me how ∑ F = 0 can be applied to a slope scenario, so i could see where the forces, their components and angles involved come from the "F=ma equation". As you say ∑ F = 0 along the normal which is ∑ F = 0 along the vertical? and the N could be found but how?

When it comes to friction problems, we could only relate work done to ΔKE if they give us a speed and ask for a distance otherwise we can not work out a force they require from the energy equations?

thanks

5. Nov 4, 2013

### tiny-tim

hi hbk69!
no, it comes from the F=ma vertical

ma = 0 (vertically), so that becomes F = 0 (ie Fg-Fsinθ = 0)
that's the horizontal equation, yes
you're confusing maths (geometry) with physics

maths (geometry) is about position (and speed and acceleration)

you're talking about the forces (which of course include the weight, mg)

the forces go on the LHS of F = ma

i'm talking about the acceleration, which goes on the RHS of F = ma … it has nothing to do with forces … you don't need to know why the acceleration is 0 … the question tells you that it is zero!

do you see that?​

6. Nov 4, 2013

### hbk69

tiny tim, how does the question tell me that the acceleration is 0?

7. Nov 4, 2013

### tiny-tim

all these three questions make it clear that the object stays on the slope

so the distance from the slope in the normal direction is always zero

so the normal acceleration is zero

are you convinced now?

8. Nov 5, 2013

### hbk69

yes sought off thanks, just confused about the earilier questions i asked in the post

9. Nov 5, 2013

### haruspex

Which ones? In the light of what you have learnt from this thread, maybe you can clarify some of them.

10. Nov 6, 2013

### hbk69

When it comes to the problems involving a slope i think i am still lost. When i use F=ma along the slop and when N,μN are involved how would i include the angles,? i know that the Normal is upwards the weight is downwards and the friction force is in the opposite direction of the objects motion but how would i relate all of these to find out what the question asks me from the F=ma. I give you an example, i was reading a book on inclined planes and i found a number of equations but did not understand them properly:

Fpull-Fpush= ma

Fpush=mgsinθ+Ff(friction), where Ff=μsFn, where Fn=mgcosθ

Fpull=mgsinθ+usmgcosθ

Facceleration=mgsinθ - μkmgcosθ

What is the Fpull/Fpush and the associated variables such as mgsinθ, cos etc must be the different forces acting on the relevant object. I want to be able to read the situation correctly and automatically know how to derive the correct formula from F=ma when it come to the slopes

Also could you show me how ∑ F = 0 can be applied to a slope scenario, so i could see where the forces, their components and angles involved come from the "F=ma equation". As you say ∑ F = 0 along the normal which is ∑ F = 0 along the vertical? and the N could be found but how?

When it comes to friction problems, we could only relate work done to ΔKE if they give us a speed and ask for a distance otherwise we can not work out a force they require from the energy equations?

11. Nov 6, 2013

### haruspex

Not straight up. It is is perpendicular to the slope.
'pull' and 'push' don't seem very descriptive to me out of context. Maybe they were more meaningful in the context of the problem.
I hope you do not think of that as a formula to be remembered. It will not always be right. I can tell from that equation that an object is being pushed up a slope, and this is opposed by both gravity and friction. Gravity and friction do not act in the same direction here, so the force of gravity is split into two parts: its component down the slope, mg sin(θ), and its component normal to the slope, mg cos(θ). (Do you understand how to resolve forces like that?)
The down slope component is in the same direction as the friction, so you can now add their magnitudes: mgsinθ+Ff.
But suppose the problem were about dragging the object down the slope. Now the component of gravity down the slope and the friction would act in opposite directions and you would be subtracting, not adding.
I note you have written μs. That implies there is nothing moving yet, and the question is about the force necessary to get the object to start moving up the slope.
This is the same as the previous equation but with pull instead of push. No real difference.
We have our old friend mgsinθ, so that's the component of gravity down the slope. Here we see it being opposed by friction (the subtraction), and its kinetic friction, so the object is moving. The equation implies these are the only two forces acting parallel to the slope, so the question asks how fast the object will accelerate down the slope with only friction to slow it.
Practise.
No, the normal is not necessarily the vertical. The normal to a slope is the line at right angles to it. In any problem about slopes you have a choice about how to resolve forces. You can resolve in the vertical and horizontal, or along the slope and normal to the slope. Either way it's two directions at right angles. Along the slope and normal to the slope often works out more simply, but either will do it.
Given the frictional force and the distance over which it acts you can work out the work done against it. You don't need the speed.

12. Nov 8, 2013

### hbk69

thanks for being patient with all my questions and helping me. Have helped my understanding a great deal.

13. Nov 9, 2013

### hbk69

so the force of gravity is split into two parts: its component down the slope, mg sin(θ), and its component normal to the slope, mg cos(θ). (Do you understand how to resolve forces like that?)

In any problem about slopes you have a choice about how to resolve forces. You can resolve in the vertical and horizontal, or along the slope and normal to the slope

^^I do not understand the highlighted above, how to resolve forces the way you have asked above could you show me how please. Thanks again, had glanced through your post earlier just read it thoroughly now

14. Nov 9, 2013

### tiny-tim

hi hbk69!

forces are vectors, and they add like vectors

now, using that vector as hypotenuse, draw any right-angled triangle on it

if the angles of that triangle are θ and 90° - θ, then the lengths of the side of the triangle are Fcos and Fsinθ

and so Fcosθ and Fsinθ in those two perpendicular directions add up to F

in other words: any vector F can be resolved as the sum of two forces in any two perpendicular directions (and the magnitudes will be Fcosθ and Fsinθ) …

and we usually choose horizontal-and-vertical or normal-and-parallel because those happen to be most convenient!