How to calculate Apparent Weight of a passenger in a car going over a bump?

[Mohmmad Maaitah]

Homework Statement

A car traveling on a straight road at 9.0 m/s goes over a hump in the road. The hump may be regarded as an arc of a circle of radius 11.0 m. (a) What is the appar- ent weight of a 600-N woman in the car as she rides over the hump? (b) What must be the speed of the car over the hump if she is to experience weightlessness? (The apparent weight must be zero.)
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How to calculate Apparent weight and did i analysis is it right?
Would the woman fell less weight on the top of the hump?

Relevant Equations

I used for (a) W-N=(m)(v^2)/r

r

 

[Mohmmad Maaitah]

Is the normal force the apparent weight?
So i just calculate it or am i mistaking

 

[BvU]

Hi, can you post your work? I only see unexplained screen shots with numbers popping up out of the blue....

 

[BvU]

(I think you may be doing ok, but the woman will be very angry at being taken for a car)

Also, there is an underlying assumption about the size of the car...

$\$

 

[haruspex]

Is the normal force the apparent weight?

Yes, and I agree with your answers.
There is one flaw with the question, though. If the top of the hill is like an arc of a circle then the vehicle is more likely to become airborne before it reaches the peak than when at the peak.

 

[Mohmmad Maaitah]

 

[Mohmmad Maaitah]

(I think you may be doing ok, but the woman will be very angry at being taken for a car)

Also, there is an underlying assumption about the size of the car...

$\$

Do you mean i have a mistake in masses?

 

[Mohmmad Maaitah]

Yes, and I agree with your answers.
There is one flaw with the question, though. If the top of the hill is like an arc of a circle then the vehicle is more likely to become airborne before it reaches the peak than when at the peak.

So the apparent weight = Normal force?
In this problem i mean

 

[haruspex]

View attachment 325563

That's odd… your first answer of 150N was closer.
You should keep more intermediate significant figures than you present as answer, e.g. 61.22, not 61.2.

 

[kuruman]

Yes, and I agree with your answers.
There is one flaw with the question, though. If the top of the hill is like an arc of a circle then the vehicle is more likely to become airborne before it reaches the peak than when at the peak.

The condition for staying on the surface is $v^2\leq gR\sin\!\theta$ where $\theta$ is measured relative to the horizontal. For the given speed of 9.0 m/s, the surface of the bump must rise at initial angle $\varphi \geq \arcsin (\frac{v^2}{gR}) = 48.7^{\circ}$ with the horizontal. At that angle, the (point) car traveling at 9.0 m/s will just barely keep contact with the surface. So if the 9.0 m/s speed is maintained until the car reaches the top, the car will not become airborne. The problem does not give the angle subtended by the arc, so we have to assume that it will be OK.

 

[haruspex]

The condition for staying on the surface is $v^2\leq gR\sin\!\theta$ where $\theta$ is measured relative to the horizontal. For the given speed of 9.0 m/s, the surface of the bump must rise at initial angle $\varphi \geq \arcsin (\frac{v^2}{gR}) = 48.7^{\circ}$ with the horizontal. At that angle, the (point) car traveling at 9.0 m/s will just barely keep contact with the surface. So if the 9.0 m/s speed is maintained until the car reaches the top, the car will not become airborne. The problem does not give the angle subtended by the arc, so we have to assume that it will be OK.

That works for (a), but, as I should have clarified, my objection is to part (b). For any $\phi>0$, the required speed will be correspondingly less than the given answer.
(And did you mean initial angle $\varphi \leq \arcsin (\frac{v^2}{gR})$ with the horizontal?)

 

[kuruman]

(And did you mean initial angle $\varphi \leq \arcsin (\frac{v^2}{gR})$ with the horizontal?)

Yes I did. I gave a different name to the complementary angle of $\theta$ but forgot to flip the inequality. Thanks.