How Is Maximum Car Acceleration Calculated with a Friction Coefficient of 0.30?

AI Thread Summary
The maximum acceleration of a car with a coefficient of static friction of 0.30 can be calculated using the formula Fp = (mew)s * mg, where Fp is the force due to friction, mew is the coefficient of static friction, m is the mass of the car, and g is the acceleration due to gravity. The user initially struggled with the problem, expressing confusion over the need for the car's mass or initial velocity. However, they later realized that the solution is straightforward and involves simply multiplying the friction coefficient by the acceleration due to gravity. This results in the maximum acceleration being 0.30 times the gravitational acceleration. The discussion highlights the importance of understanding the relationship between friction and acceleration in vehicle dynamics.
tbaskerv
Messages
7
Reaction score
0

Homework Statement




What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.30?

Homework Equations


sigmaFy = Fn-mg = may = 0
sigmaFx = Fp - Ffr = max = 0
Ffr = (mew)s * Fn
Fp = Ffr = (mew)*Fn
Fp = (mew)s(mg)

mg is mass times gravity
* are multiplication
(mew)s = static friction
Fn = normal force


The Attempt at a Solution



Um...I have no clue how to solve it, any direction to point me in would be helpful. I can't see how to solve it without the mass of the car, or the initial velocity.
 
Physics news on Phys.org
Nevermind, it was as simple as multiplying the coefficient by gravity, sorry for the space waster.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Banked roads -- circular motion and gravitation
Replies
11
Views
2K
Coefficient of Friction question for a cart going down a wooden ramp
Replies
18
Views
3K
Centripetal force problem involving a washing machine
Replies
5
Views
3K
How Do You Calculate the Weight of a Block on an Inclined Plane?
Replies
6
Views
3K
Newtons Laws car deacceleration
Replies
4
Views
5K
How Does Static Friction Determine a Car's Maximum Acceleration?
Replies
2
Views
6K
Angular velocity- A toy train rolls around a horizontal track
Replies
8
Views
752
Coefficient of Friction when Normal Force is Reduced [HS]
Replies
5
Views
2K
Finding friction coefficient, car unbanked curve.
Replies
3
Views
3K
Static Friction Between a Box and the Floor
Replies
11
Views
3K

Hot Threads

Recent Insights

Back
Top