Finding the friction coefficient?

AI Thread Summary
To determine the minimum coefficient of static friction required for a Porsche to accelerate at 11.2 m/s² without spinning its tires, the net force in the horizontal direction must be analyzed using Newton's second law. The equation for static friction (fs) is related to the normal force (N) and the coefficient of static friction (μs) as fs = μs * N. Although mass (m) is not provided, the acceleration (a) is given, allowing for the calculation of the necessary frictional force to prevent tire spin. By applying the equation F_net = ma and recognizing that the normal force equals the weight of the vehicle (N = mg), the coefficient of static friction can be derived without needing the mass explicitly. This approach clarifies that the key to solving the problem lies in understanding the relationship between force, mass, and acceleration.
papi
Messages
31
Reaction score
0

Homework Statement



Hopping into your Porsche, you floor it and accelerate at 11.2 m/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.

Homework Equations




well I know fs/N= the coefficient, however, even though N=ma, you dnt have an m nor an a. i just dnt see adequate info given. help!
 
Physics news on Phys.org
papi said:

Homework Statement



Hopping into your Porsche, you floor it and accelerate at 11.2 m/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.

Homework Equations




well I know fs/N= the coefficient,
yes
however, even though N=ma,
why do you say N=ma?
you dnt have an m nor an a. i just dnt see adequate info given. help!
the acceleration, a, is given. Write the Newton 2 equation, looking in the x direction. What's the net force in that direction? Maybe you won't need to know m
 

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

Static Friction Between a Box and the Floor
Replies
11
Views
3K
How to find coefficient of friction *Without* mass?
Replies
3
Views
22K
How to find coefficient of friction of a body sliding down the slope?
Replies
7
Views
2K
How to find the coefficient of kinetic friction
Replies
4
Views
3K
Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Find the force of Friction and minimum coefficient of friction
Replies
1
Views
2K
Coefficient of friction of a sandwiched body
Replies
15
Views
799
How can I determine the instant the block overcomes static friction?
2
Replies
61
Views
3K
Coefficient of friction between puck and ice
Replies
2
Views
3K
Finding Force by using the coefficient of friction and mass
Replies
16
Views
3K

Hot Threads

Recent Insights

Back
Top