# physics calculating force of friction

Thread Starter

#### zelda1850

Joined Jan 3, 2010
18
A force of 10 newtons toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden crate. the crate weights 25 newtons.

1)caculate the magnuitude of the force of the friction between the crate and floor

equation is u*fn = ff

i know the applied force is 10 n and it is exerted so there is static energy how can i find u and fn?

2) what is the magnitude of the net force acting on the crate

net force = ma

i cant tell the acceleration or mass how can i find fn?

#### Fraser_Integration

Joined Nov 28, 2009
142
What is u*fn = ff meant to be?

To get friction force you need to know the co-efficient of friction, u (mu), and then F = uR, where R is the normal contact force, which will be equal to the weight in this case.

Thread Starter

#### zelda1850

Joined Jan 3, 2010
18
so the normal force is 25 newtons but how can i find mu?

#### studiot

Joined Nov 9, 2007
4,998
A force of 10 newtons toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden crate. the crate weights 25 newtons.
I think you should practise reading the question more carefully.
However I think I know what they are getting at.

1) The frictional force is variable and always exactly equal (but opposite) to the 'pushing' force until motion just starts. Since we are talking horizontal and vertical forces here this part is about the horizontal forces and I guess that your question actually says the crate just starts to move with a 10KN horizontal force.

2) Once the crate starts to move the frictional force does not increase, merely accelerate with any additional 'pushing' force. Since are told the weight of the crate we also know the vertical forces involved. Remember that the weight acts downwards on the support. The Reaction acts upwards on the crate.

So your answer will be the (vector) combination of the rightwards horizontal push, the leftwards horizontal friction and the upwards reaction.

I strongly suggest you draw a diagram.

Without further information this question is not about mu or acceleration.

Thread Starter

#### zelda1850

Joined Jan 3, 2010
18
so the horizontal force on the crate should be 10 n and that is the app force?
and the vertical forces is 25n so that is fnormal and fg force?

so how can i find the caculation of the fritional force knowing the horizontal and vertical forces?

#### studiot

Joined Nov 9, 2007
4,998
so the horizontal force on the crate should be 10 n and that is the app force?
and the vertical forces is 25n so that is fnormal and fg force?
I didn't say either of those things.

Have you drawn a diagram and can you post it?

Thread Starter

#### zelda1850

Joined Jan 3, 2010
18
emm i drew the diagram with the vertical forces of fn and fg as 25n and the horizontal force 10 n

#### studiot

Joined Nov 9, 2007
4,998
What are fn and fg?

What about the frictional force, you haven't included it?

Thread Starter

#### zelda1850

Joined Jan 3, 2010
18
but how can i find the frictional force with the info i have? i thought it was mu*fn but it so confusing now

#### studiot

Joined Nov 9, 2007
4,998
OK

Carefully reread my post #4

Is the crate flying away - no - it is is vertical equilibrium.
The reaction from the floor is exactly enough to support the crate. NO more NO less.
There are two vertcial forces acting on the crate, the weight (downwards) and the reaction (upwards) . And they are in balance.

If I place a weight on top of the crate, the reaction increases to exactly match the additional weight.

Reaction = Weight

R = W

Now Friction works in exactly the same way as I said in post 4.

But the friction foce is parallel to the contact surface (floor).

The reaction force is at right angles to it (normal).

Thus the friction force (F) is always at right angles to the reaction force.

So in your case the reaction is vertical so the friction force (F) is horizontal.

Let us now imagine the 10N push to the right being slowly applied.

As it increases from zero through 1N, 2N, 3N etc the crate does not move.
This is another word for equilibrium.

So the frictional force exactly matches and opposes the applied push.

Until you get to 10N.
This is the maximum you can achieve with this crate.

If you push even the smallest bit harder than 10N the crate will move.

At this point the frictional force equals (but opposes)the push of 10N and is horizontal.
So there are two horizontal forces acting on the crate. The push (to the right) and the frictional force (to the left). And they are in balance up to 10N.

You do not need the coeffiecient of friction (μ) although you can calculate it if required.

(μ) = F/R = 10/25

Last edited:

#### schmity

Joined Feb 3, 2010
1
Just another thought...If there is no acceleration while the crate is subjected to the force of 10 (i.e. the crate stays moving at the same velocity), mu would simply be 25/10 and would be a kinetic friction coefficient.

#### studiot

Joined Nov 9, 2007
4,998
Just another thought...If there is no acceleration while the crate is subjected to the force of 10 (i.e. the crate stays moving at the same velocity), mu would simply be 25/10 and would be a kinetic friction coefficient.
That's exactly right.

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