Calculating Friction Force on 5000kg Truck on 14° Slope

[aligass2004]

Homework Statement

A 5000 kg truck is parked on a 14 degree slope. How big is the friction force on the truck?

Homework Equations

friction = coefficient of friction X normal force

The Attempt at a Solution

I tried breaking the weight into components. I used 1.00 as the coefficient of friction (given in a table in the book...rubber on concrete). I got 60916.27 N, which of course was wrong.

 

[Dick]

You don't need no stinking tables. The force of friction acting on the truck is exactly equal and opposite to the force attempting to drag the truck down the slope. Since the truck is 'parked' and not moving, total force on it is zero.

 

[aligass2004]

Zero isn't right.

 

[Dick]

Of course it isn't. Total force is tangential gravitational force plus frictional force. They act in opposite directions. What is the component of gravitational force acting down the incline? That's equal to the frictional force.

 

[aligass2004]

The gravitational force down the incline is mgcos(theta), which equals 47593.005. I tried that as well.

 

[Dick]

I think the normal force is mg*cos(theta). I think the downward tangential force is mg*sin(theta).

 

[aligass2004]

That was right. Thank you!

 

[Dick]

That was right. Thank you!

I hope you know why that was right. But you're welcome.

 

[pooface]

Take a look at my thread for a similar problem called "Friction force problem".

You are almost there to solving this. mgcos(14) gives you the normal force. which is perpendicular to the slope. It should also be in mega Newtons.

Look at your formula.

The answer I am getting is 11.866 MegaN. I am studying the samething so. yea.

EDIT: damn i was too late.

 

[Dick]

Take a look at my thread for a similar problem called "Friction force problem".

You are almost there to solving this. mgcos(14) gives you the normal force. which is perpendicular to the slope. It should also be in mega Newtons.

Look at your formula.

The answer I am getting is 11.866 MegaN. I am studying the samething so. yea.

EDIT: damn i was too late.

You are not only late, the answer you gave is pretty wrong. Suggest you figure out why.

 

[pooface]

You are not only late, the answer you gave is pretty wrong. Suggest you figure out why.

Ok. I re-did the problem and now am getting 11.8649 kN as the frictional force.

For some reason I converted the 5000 kg to grams and then did multiplied by 9.81 m/s^2.

Is that the correct answer?

The static friction coeff = tan14 = 0.2493
FN = 47593 N

(0.2493 x FN)= FF

= 11.8649 kN

 

[aligass2004]

I got 11866.27 N