Two Cars Coming to Rest on an Incline

[Jennings]

Homework Statement

A car weighing 1940lb has lost its brakes and is unable to come to a stop. A police cruiser of 3900lb gets ahead of it so that the two cars are now bumper to bumper. The combined cars are now traveling at 60mph down a 15 degree grade slope. If the police cruiser can normally brake from 60mph to rest in 250ft if it is on level ground:

A) How long will it take for the police cruiser to bring both cars to a stop?

B) How far will they have traveled during this time?

Ignore friction. [/B]

Homework Equations

[/B]
F= ma
V^2 - Vo^2 = 2a∆x

The Attempt at a Solution

The first thing i did was convert everything into SI units for simplicity.
Speed of Cars
26.8 m/s
Mass of car
881kg
Mass of police cruiser
1770kg
Weight of car
8643N
Weight of police cruiser
17360N
Weight of Combined Cars
26000N

I then determined the breaking force Fb of the police cruiser's brakes by determining first the acceleration of caused by the brakes, and then finding the force using F=ma
[V^2 - Vo^2] / [2∆x] = a
ab= [0 - (26.8m/s)^2]/[2(76.3m)] = -4.71m/s^2

Fb=mab
F=(1770kg)(-4.71m/s^2) = -8337N

So the force applied by the brakes of the police is -8337N

The next thing i did was draw a free body diagram of the cars going down a 15 degree slope.
So if the cars are currently moving at a speed of 26.8m/s down the slope, this means they are not accelerating? right? I am confused about what to do next .

∑Fx = Wx - Fbx ??

 

[haruspex]

According to your FBD, what is the net force on the two car system, measured along the slope?
But there is a catch with this problem you may have overlooked. It doesn't tell you what the limiting factor is on how quickly the police cruiser can stop. Is it the brakes themselves or the grip on the road? Assuming that fully applied the brakes could cause the vehicle to skid, it would be the grip on the road. If so, what complication arises when the cruiser is going downhill?

 

[Jennings]

According to your FBD, what is the net force on the two car system, measured along the slope?
But there is a catch with this problem you may have overlooked. It doesn't tell you what the limiting factor is on how quickly the police cruiser can stop. Is it the brakes themselves or the grip on the road? Assuming that fully applied the brakes could cause the vehicle to skid, it would be the grip on the road. If so, what complication arises when the cruiser is going downhill?

The forces in the direction of the slope would be the component of the weight and the breaking force .

so
∑Fx = Wx - 8337N

And Wx is just Wsinθ which is

 

[Jennings]

According to your FBD, what is the net force on the two car system, measured along the slope?
But there is a catch with this problem you may have overlooked. It doesn't tell you what the limiting factor is on how quickly the police cruiser can stop. Is it the brakes themselves or the grip on the road? Assuming that fully applied the brakes could cause the vehicle to skid, it would be the grip on the road. If so, what complication arises when the cruiser is going downhill?

A

The forces in the direction of the slope would be the component of the weight and the breaking force .

so
∑Fx = Wx - 8337N

And Wx is just Wsinθ which is

6720N

∑Fx = 6720N - 8337N= -1617N

atleast to my knowledge this is what it would be.

Now as far as this complication i have no idea. If it is something advanced than it is probably meant to be disregarded.. I am taking physics 1

 

[PeroK]

Homework Statement

A car weighing 1940lb has lost its brakes and is unable to come to a stop. A police cruiser of 3900lb gets ahead of it so that the two cars are now bumper to bumper. The combined cars are now traveling at 60mph down a 15 degree grade slope. If the police cruiser can normally brake from 60mph to rest in 250ft if it is on level ground:

A) How long will it take for the police cruiser to bring both cars to a stop?

B) How far will they have traveled during this time?

Ignore friction. [/B]

Homework Equations

[/B]
F= ma
V^2 - Vo^2 = 2a∆x

The Attempt at a Solution

The first thing i did was convert everything into SI units for simplicity.
Speed of Cars
26.8 m/s
Mass of car
881kg
Mass of police cruiser
1770kg
Weight of car
8643N
Weight of police cruiser
17360N
Weight of Combined Cars
26000N

I then determined the breaking force Fb of the police cruiser's brakes by determining first the acceleration of caused by the brakes, and then finding the force using F=ma
[V^2 - Vo^2] / [2∆x] = a
ab= [0 - (26.8m/s)^2]/[2(76.3m)] = -4.71m/s^2

Fb=mab
F=(1770kg)(-4.71m/s^2) = -8337N

So the force applied by the brakes of the police is -8337N

The next thing i did was draw a free body diagram of the cars going down a 15 degree slope.
So if the cars are currently moving at a speed of 26.8m/s down the slope, this means they are not accelerating? right? I am confused about what to do next .

∑Fx = Wx - Fbx ??

Yes, on a downhill slope, you have the normal breaking force less the gravitational force down the slope.

Although there is nothing wrong, as such, with plugging in the numbers as soon as possible, you might consider solving the problem formulaically and then plugging in the numbers at the end. This has the advantage of showing you the relationship between everything, rather than a sequence of numerical values.

 

[PeroK]

A6720N

∑Fx = 6720N - 8337N= -1617N

atleast to my knowledge this is what it would be.

Now as far as this complication i have no idea. If it is something advanced than it is probably meant to be disregarded.. I am taking physics 1

Yes, that's right.

The complication is that the maximum braking force probably depends on grip on the road and when going downhill there is a lower normal gravitational force, hence less grip, hence less maximum braking force. But, you can safely ignore this.

 

[haruspex]

Yes, that's right.

The complication is that the maximum braking force probably depends on grip on the road and when going downhill there is a lower normal gravitational force, hence less grip, hence less maximum braking force. But, you can safely ignore this.

I don't know why you say it can be ignored. The OP clearly states that the given info on braking capacity is for a level road. If the questioner intends that the reduction in normal force should be ignored then I doubt this would have been mentioned.

 

[haruspex]

Although there is nothing wrong, as such, with plugging in the numbers as soon as possible, you might consider solving the problem formulaically and then plugging in the numbers at the end. This has the advantage of showing you the relationship between everything, rather than a sequence of numerical values.

I would put it rather more strongly. It is a very good idea, for many reasons, to delay plugging in numbers until the final step. Not least, it makes it much easier for others to follow the logic.

 

[PeroK]

I don't know why you say it can be ignored. The OP clearly states that the given info on braking capacity is for a level road. If the questioner intends that the reduction in normal force should be ignored then I doubt this would have been mentioned.

Only the question setter really knows. But, without an assumption that grip is a limiting factor, there's no way to calculate braking force at an angle. Given the level of the problem, I'd assume that it's the brakes themselves that are the limiting factor.

 

[haruspex]

. But, without an assumption that grip is a limiting factor, there's no way to calculate braking force at an angle. .

Ok, but it is a very reasonable assumption. If I brake violently (in the absence of ABS) I expect the tyres to skid on the road, not the brake pads to skid on the drum. I think it most likely that the question setter overlooked the need to state that. The proviso of a level road is strong evidence that the reduction in normal force should be allowed for.