PHY problem

[joonkimdds]

"Suppose that u r standing on a train accelerating at 0.24g. what minimum coefficient of static friction must exist between ur feet and the floor if u r not to slide down?"

could someone tell me how i should set up this problem?

---

Members don't see this ad.

 

[Sicilian]

"Suppose that u r standing on a train accelerating at 0.24g. what minimum coefficient of static friction must exist between ur feet and the floor if u r not to slide down?"

could someone tell me how i should set up this problem?

F = (coefficient of static friction)(Fn), where F is the static friction force and Fn is the normal force.

Any force is equal to ma. So rewrite the above as:

ma = (coefficient of static friction)(m)(g), where the normal force, Fn, is equal to the weight of the person.

So the coefficient of static friction = a/g = (2.352) / (9.8) = 0.24, which is a dimensionless quantity.

 

[joonkimdds]

F = (coefficient of static friction)(Fn), where F is the static friction force and Fn is the normal force.

Any force is equal to ma. So rewrite the above as:

ma = (coefficient of static friction)(m)(g), where the normal force, Fn, is equal to the weight of the person.

So the coefficient of static friction = a/g = (2.352) / (9.8) = 0.24, which is a dimensionless quantity.

well, first of all, u definitely got the answer right 🙂
but i also got a question for ya.
when i see a problem like this, i tend to draw a picture with arrows incicating different forces.

so in this problem , i drew a person on the train.
Then there r FN arrow up, MG arrow down, Ff to the right(since that's what keeps the person from going to left), and then finally force that pushes both the person and the train(i don't know if it's left, right or not even exist).

FN and MG cancels out each other.
So what do did was Fnetx = MAx.
Ff+Fpush = MA
Ff = MA - Fpush

i don't know if i am talking about a nonsense but could u tell me where i went wrong?

 

[Sicilian]

well, first of all, u definitely got the answer right 🙂
but i also got a question for ya.
when i see a problem like this, i tend to draw a picture with arrows incicating different forces.

so in this problem , i drew a person on the train.
Then there r FN arrow up, MG arrow down, Ff to the right(since that's what keeps the person from going to left), and then finally force that pushes both the person and the train(i don't know if it's left, right or not even exist).

FN and MG cancels out each other.
So what do did was Fnetx = MAx.
Ff+Fpush = MA
Ff = MA - Fpush

i don't know if i am talking about a nonsense but could u tell me where i went wrong?

The person is in a stationary position, relative to the floor. Since he isn't moving, there is no acceleration:

Fx = 0

Fy = 0

Even though the person isn't moving, the static friction force still takes into account the normal force (Fn). If you assume that the person's shoes are in contact with the floor, and the person is standing still, his shoes have actually "penetrated" the floor. You can't see it on a macroscopic scale, but no surface is 100% smooth (not even ice). So if you brought two surfaces in contact with each other, they would actually "fit" onto each other, kind of like the way a lid fits onto a container. The static friction force measures the degree of the penetration, when one surface is at rest on top of the other surface. The degree of penetration varies from surface to surface (this is why you have coefficients of friction). In equation form:

Static friction force = coeff of static fric X normal force

So anytime you have friction, you're still dealing with a normal force. Thats what makes it possible to solve this problem.