Nero25
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This problem is driving me crazy!
What is the magnitude of the force of the car on the truck, when the mass of the car is 1000 kg, when the mass of the truck is 2000 kg, when the driver steps on the accelerator, the wheels of the car push against the ground with a force of 4500 Newtons. What is the magnitude of the force of the car on the truck? The car is pushing the truck, just to clarify, not the other way around.
The car is pushing the truck, there's no rope or chain or anything involved. It seems like there's some sort of information missing in this problem.
F = m x a
Force / Normal force = coefficient of friction
I started off by assuming you could find the force by multplying the mass times the acceleration, because of the fact that the problem gives you so little information, but that's incorrect. (1000 x 9.807)
I do know that all the forces given are in the y-direction, and there's no way to change that to x without finding the acceleration or the coeffecient of friction. I'm not quite sure if you can use the Newtons as the force to find a coefficient of friction, in which case I got 4500/9800, which is equal to .46 for the frictional coefficient, but I'm not sure if that's right or not.
Someone please help!
Homework Statement
What is the magnitude of the force of the car on the truck, when the mass of the car is 1000 kg, when the mass of the truck is 2000 kg, when the driver steps on the accelerator, the wheels of the car push against the ground with a force of 4500 Newtons. What is the magnitude of the force of the car on the truck? The car is pushing the truck, just to clarify, not the other way around.
The car is pushing the truck, there's no rope or chain or anything involved. It seems like there's some sort of information missing in this problem.
Homework Equations
F = m x a
Force / Normal force = coefficient of friction
The Attempt at a Solution
I started off by assuming you could find the force by multplying the mass times the acceleration, because of the fact that the problem gives you so little information, but that's incorrect. (1000 x 9.807)
I do know that all the forces given are in the y-direction, and there's no way to change that to x without finding the acceleration or the coeffecient of friction. I'm not quite sure if you can use the Newtons as the force to find a coefficient of friction, in which case I got 4500/9800, which is equal to .46 for the frictional coefficient, but I'm not sure if that's right or not.
Someone please help!