# Why can this question about Centripetal Force not be answered in this way? [closed]

Here is the question:

"A car travels round a bend which has radius $100~\text{m}$ and is banked at an angle of $20°$ to the horizontal. The car is travelling at a speed of $30 ~\text{m}\text{s}^{-1}$. What is the least possible value of the coefficient of friction if the car does not slip up the slope?"

The way my textbook says to answer it is to resolve the force of gravity and the friction $(F=μR)$ in the vertical and horizontal directions and to consider centripetal force to be completely horizontal.

Solving in this way will give you two equations (one from the vertical component and one from the horizontal component) which you can solve simultaneously to give the solution:

$μ = 0.416$

However, I tried answering it by instead resolving the centripetal force in the direction of the slope and then equating this component to the friction and the force of gravity down the slope as follows:

$F_\text{centripetal} = \frac{m(30^2)}{100} = 9~\text{m}$

$F_\text{gravity} = mg$

then equating the component of centripetal force down the slope to the friction and the component of gravity down the slope (where I have used F=μR for friction):

$9m\cos(20°) = mg\sin(20°) + μmg\cos(20°)$

Giving the solution $μ = 0.554$

I used the same value for gravity as they did in their answer: $g = 9.8~\text{m}\text{s}^{-2}$

Could somebody please tell me why there is a difference in the answers? Thanks!

## closed as off-topic by ACuriousMind♦, John Rennie, Ryan Unger, HDE 226868, JimAug 27 '15 at 16:46

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• In your equation $F_\text{cent} \cos(20^\circ) = mg \sin (20^\circ) + \mu mg \cos(20^\circ)$, are you making the assumption that the normal force exerted by the track is equal to $mg$? If you are, you shouldn't. – Michael Seifert May 25 '15 at 20:04
• I don't think I did. $mg\cos(20∘)$ is the normal reaction force to the slope and then since the frictional force $F$ is given by $F=μR$, I simply multiplied the above by $μ$ to get the frictional force opposing the centripetal force down the slope. The equation can be read as "The component of the centripetal force up the slope (the force required to keep the car in the same position at said velocity of 30ms^-1) is equal to the force of gravity down the slope plus the frictional force down the slope" for clarification. – Resquiens May 25 '15 at 20:27

At the risk of sounding like a broken record... it is a good idea to draw a diagram for all but the very simplest problems (and even then):

You can immediately see that the normal force is made up of two components: $F_c \sin\alpha$ and $F_g\cos\alpha$. The friction results from the combination of both of these.

In your approach, you ignore the normal force due to centripetal component. Which is why you get the wrong answer ($\mu$ too large since the normal force you used was too small).

The problem is that the equation should be $9m\cos 20= mg \sin 20+\mu mg$ since the friction force is going along the slope.

• Can you explain? Surely you need to include the $\cos20$ in the expression for friction because it is proportional to the normal reaction force which is equal to $mg\cos20$. Also, using your equation yields the result $μ=0.52$, which is still not the same anyway. – Resquiens May 26 '15 at 16:43
• Oh yeah, you are right. Then I don't know what to do. – Iván Mauricio Burbano May 26 '15 at 16:46
• OK, thanks for your effort though - if you do find inspiration in future please come back and share :) – Resquiens May 26 '15 at 16:47

If we use up as $y$ and left as $x$ then the horizontal components of the Normal and Friction force are what supply the centripetal force. Hence: $$N\sin{20^{\circ}} + f\cos{20^{\circ}} = 9m$$ Evaluating vertical forces as well we can get the next expression: $$N\cos{20^{\circ}} - f\sin{20^{\circ}} - mg = 0$$ Now, using the expression: $$f = u_kN$$ by substituting, rearranging and solving (using all three equations) you can solve for $u_k$. I hope this has helped.