Redbelly98 said:Try drawing a free body diagram of the car. What are the forces that are present, and in what direction are they?
Also, in what direction (horizontal or vertical) does the car decelerate? Which force or forces are acting in that direction?
Redbelly98 said:Yes: gravity downwards, normal force from ground upwards, and friction force horizontally. And two of those forces are exerted by the ground, so the car exerts equal-but-opposite forces on the ground.
kthejohnster said:a car of mass 1500kg and decelerating 15ms-2. what is the total force of car on road from all tires?
The two forces act in different directions, x or y, so we can't simply add them. But yes, they turn out to be equivalent to mg and ma -- I guess this is what you were saying in your first post. But technically you need to say in what direction those forces are acting. So for the 15,000 N force you calculated from mg, in what direction does it act? Likewise for the 22,500 N force you calculated from ma.kthejohnster said:so would that be mg+ma because opposite of gravity would be mg and the force opposite friction seems to be ma or is there a separate y force and x force
While you bring up a valid point, since this is intro physics it's probably best to keep things simple; neglect air resistance, and use the numbers given. It's likely that the question author was not thinking through about the implications of a 1.5g acceleration due to friction -- or the 15 m/s2 was a simple typo.cmb said:Difficult to tell. I expect this is not the answer the question is looking for, but it means you are in the process of having an accident so your tyres might not be generating any force at all!
I do not believe your tyres can generate 1.5g (I think the limit for road cars is around 1.1g with latest tech), unless you are in a race car with aerodynamic downforce, but then the car wouldn't weight more than 605kg. (due to Formula 1 construction rules 605kg or Formula 3 540kg.)
To calculate the force of a 1500kg car on the road, you need to use the equation F = m x a, where F represents force, m represents mass (in kilograms), and a represents acceleration (in meters per second squared). In this case, you will also need to know the acceleration of the car, which can be determined by dividing the change in velocity by the time it takes to change that velocity.
The standard unit of force used in this calculation is the Newton (N). One Newton is equal to 1 kilogram-meter per second squared (kg-m/s^2).
The weight of the car, which is the force of gravity acting on the mass of the car, does not directly affect the force on the road. However, the weight of the car does contribute to the overall mass of the car, which is a factor in calculating the force on the road.
Yes, according to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that the force of the car on the road is equal in magnitude and opposite in direction to the force of the road on the car.
The force of the car on the road is directly proportional to the car's acceleration. This means that the greater the force of the car on the road, the greater the car's acceleration will be. This is why cars with more powerful engines are able to accelerate faster than cars with weaker engines.