Circular Motion and Friction of car turning corner

[haroldtreen]

Homework Statement

"A 1500kg car rounds a curve of radius 75m at a speed of 25m/s. The curve is banked at an angle of 22 degrees to the horizontal. Calculate:

a) The magnitude of the frictional force.
b) The Coefficient of Kinetic Friction."

Homework Equations

F=mv$$^{2}$$/r

Answers given:
a) 61000N
b) 0.33

The Attempt at a Solution

What makes sense to me is that the centripetal force is countered by another force. I believe this other force is a combination of the horizontal frictional force & the horizontal normal force (cause from the banked curve resisting the centripetal force).

The centripetal force = 12500N

So,

12500N = Force Friction x + Force Normal x = Force Friction x + Force Normal*Sin$$\phi$$

Force Normal = mg/cos$$\phi$$

This however have been unable to get me the provided answers :S.
It seems like a pretty simple question and I have been playing around with it for a bit and keep ending up with 6560N for the frictional force :(.

If anyone could help it would be great...I'm studying for exams! :P

 

[mo_0820]

Now, please make the force analysis again.
How many forces act on the car? And what is the effect?

 

[inky]

Homework Statement

"A 1500kg car rounds a curve of radius 75m at a speed of 25m/s. The curve is banked at an angle of 22 degrees to the horizontal. Calculate:

a) The magnitude of the frictional force.
b) The Coefficient of Kinetic Friction."

Homework Equations

F=mv$$^{2}$$/r

Answers given:
a) 61000N
b) 0.33

The Attempt at a Solution

What makes sense to me is that the centripetal force is countered by another force. I believe this other force is a combination of the horizontal frictional force & the horizontal normal force (cause from the banked curve resisting the centripetal force).

The centripetal force = 12500N

So,

12500N = Force Friction x + Force Normal x = Force Friction x + Force Normal*Sin$$\phi$$

Force Normal = mg/cos$$\phi$$

This however have been unable to get me the provided answers :S.
It seems like a pretty simple question and I have been playing around with it for a bit and keep ending up with 6560N for the frictional force :(.

If anyone could help it would be great...I'm studying for exams! :P

Actually your answer given for frictional force is not 61000N. Could you check this ? I think anwer is 6100 N. You need to use g =9.8 ms^-2. You can get 6083.08 N. It is around 6100N.
Then if use coefficient of friction formula, you can get mu value 0.33. Could you try again?

 

[ehild]

Force Normal = mg/cos$$\phi$$

This is not true.The force of friction has both horizontal and vertical components.

By the way, the answer for the frictional force is wrong, it is rather 6100 N.

ehild

 

[inky]

Actually your answer given for frictional force is not 61000N. Could you check this ? I think anwer is 6100 N. You need to use g =9.8 ms^-2. You can get 6083.08 N. It is around 6100N.
Then if use coefficient of friction formula, you can get mu value 0.33. Could you try again?

Use summation F(x)=(mv^2)/r
summation F(y)=0
Both normal force and friction force have 2 components.