Seydlitz said:
Homework Statement
The text-book question is, calculate the push of the road on the car with a mass of 900kg traveling with a speed of 18m s-1 over a hump-backed ridge.
The Attempt at a Solution
Push of the car on the road is equal to the push of the road to the car, hence:
ƩF = Fc + W
ƩF = (900x18/50)+(900x9.81) = 9153N
I just want to confirm if my calculation and consideration is correct since I took it that the centripetal are also one direction with the weight of the car. In general I also want to confirm if we could just take Fc as ordinary force (push and pull) without any special care on them. It's because there's a similar question asking our effective weight considering the Earth centripetal force. I know intuitively that the centripetal force will cause our body to be actually accelerated away from the Earth hence reducing the gravitational attraction. Hence instead of adding the centripetal force to the weight force I subtract it.
Effective Weight = mg - Fc?
Could anyone clarify this both problems because they seemed to be contradictory?
Thank You
Joining this discussion late:
Particularly worried about your statement I have made red, above.
Gravitational attraction force is determined by the mass of the earth, the Mass of the Car and the distance from the centre of mass of the car to the centre of mass of the Earth.
Whether you are driving the car through a dip, over a hump-backed bridge or on flat ground, none of those quantities is appreciably changed, so the Gravitational Attraction is unchanged. We usually call that force the weight, and calculate its size using "mg". We also know the direction of that force [on the car] is towards the centre of the Earth - often called "down".
On flat ground, there is no vertical acceleration, so no net force. That means the Normal Reaction force is equal in size to the weight force, but is directed up. The weight force and the Normal reaction force cancel each other giving a net force of zero - and thus the net acceleration of zero. [remember this is only a vertical consideration].
If the car is driven through part of a vertical circle [through a dip or over a humped backed bridge] there will be a resulting centripetal acceleration - achieved by there being a centripetal force. That force is directed toward the centre of the circle [of the motion].
When driving through a dip, the centre of the circle is above the ground - often called up - so the net force is up.
That is achieved by the Normal Reaction force becoming
larger than mg.
When driving over the humped back bridge, the centre of the circle is below the bridge - often called down - so the net force is down.
This is achieved by the Normal Reaction force becoming
smaller than mg.
How does the Normal Reaction force manage to change size/strength? The key is in the word reaction. The surface reacts as required [or breaks trying] and supplies just as much force as is necessary.
The real problem arises when so many people refer to the force as the Normal force - leaving out that key descriptor "reaction" which is the key to explanation.
Examples:
Place a 1kg book on a table - the table will push up with a force of 9.8 N , balancing the weight of the book - thus supporting the book.
Place a 9 kg mass on top of the book: now the table must push up with a force of 98N to balance the weight of the couple.
Add another 10kg - the table must push up with 196N
Add another 10kg - the table must push up with 294N
Keep doing this and we will reach a point when to support the book & masses, we are requiring the table to push up with a force larger than is possible. The table fails - though most people simply say the table breaks under the load.
