Tangential Acceleration and Gravity in a moving car

[Conservation]

Homework Statement

A car is driven at a constant speed of 10.0 m/s as it travels over a circular-shaped arch. If the radius of curvature of the bridge is 50.0m, what upward force does the car exert on the 64.0 kg person riding in the car as it passes over the bridge?

Homework Equations

v²/r=a_c
F_{man on car}=F_{car on man}
F_g=ma_g

The Attempt at a Solution

The centripetal acceleration is towards the center, or down towards the bottom of the arched bridge. Similarly, the man's weight is exerted downwards. Both of these forces contribute to the F_{man on car}.

Hence, F_{man on car}=ma_g+mv²/r, making the answer 755.84 N. However, the answer key says that the ma_g and the mv²/r are in different directions, making the answer 499.84 N instead. Any help on this would be greatly appreciated.

 

[jackarms]

They are in fact in different directions, so I don't know what the solution means. But for the answer make a free-body diagram.

[itex]∑F_{y} = F - mg = -ma[/itex]

So [itex]F[/itex] in fact equals [itex]mg - ma[/itex]

The error in your reasoning is that the quantity mv^2/r, the centripetal acceleration, is not provided by a separate force -- the weight force provides it, so the relationship changes. Also, it's not tangential acceleration, but instead centripetal acceleration, because it points inward.

 

[Conservation]

Wow, my apologies. I meant to say centripetal-I have no idea why I said tangential.

I don't understand what you mean by the fact that the weight force "provides the centripetal acceleration." Could you elaborate on the topic?

Also, why are they different directions? Aren't both gravitational weight and centripetal acceleration towards the center, or below the bridge?

Thank you.

 

[jackarms]

No problem -- I do it all the time.

Think about it in terms of Newton's second: [itex]F = ma[/itex]

Essentially this means that forces provide accelerations, and accelerations need forces to produce them. So the centripetal acceleration needs a force to produce it, and this is the weight force. Centripetal forces are what provide centripetal accelerations, so in this case weight force also acts as the centripetal force. Mathematically, the quantities go on separate sides of the equation, which is how you get that negative sign in the answer. That's why even though they're in the same direction, their magnitudes don't add -- they're different quantities.

EDIT: Sorry, I meant in my first post to say that weight force and acceleration are in the same direction -- just mathematically they end up opposite one another.

 

[HalfLostAndSearching]

I would say that the answer key is correct. The centripetal acceleration and the gravitational force are indeed in different directions, as one is towards the center of the circular path and the other is towards the ground. Therefore, they cannot be added together as they are not acting in the same direction.

To calculate the upward force exerted on the person, we can use the equation Fnet = ma, where Fnet is the net force acting on the person, m is the person's mass, and a is the net acceleration. In this case, the net acceleration is the sum of the tangential acceleration (v^2/r) and the centripetal acceleration (g), but we must take into account their different directions.

Therefore, the net acceleration can be calculated as √(v^2/r)^2 + g^2, which is equal to √(100/50)^2 + (9.8)^2 = 10.09 m/s^2.

Plugging this into the equation Fnet = ma, we get Fnet = (64 kg)(10.09 m/s^2) = 646.76 N.

Since the net force is equal to the sum of the upward force (Fman on car) and the downward force (Fg), we can rearrange the equation to solve for Fman on car.

Fman on car = Fnet - Fg = 646.76 N - (64 kg)(9.8 m/s^2) = 499.84 N.

Therefore, the upward force exerted on the person riding in the car is 499.84 N, as stated in the answer key.