No, it's accelerating.spartacus said:
Sorry, i meant that as a relevant equation in one dimension, since it's not accelerating verticallyDoc Al said:
Sure, the sum of the vertical force components must be zero.spartacus said:
ya so could you help me see what i did wrong?Doc Al said:
I figured N*sinθ is equal to centripetal force so it would by definition be going in the opposite direction of frictionJazz said:
For one thing, you included friction in your analysis. For another, you failed to consider the direction that friction acts.spartacus said:
In the absence of friction.spartacus said:
What's the direction of the centripetal acceleration? What's the direction of friction with respect to the surface on which it acts?spartacus said:
Why wouldn't i include friction?Doc Al said:
Because it asks for the angle of 'maximum control'. Is this question a part of a multi-part question?spartacus said:
Are you assuming that friction acts horizontally?spartacus said:
I believe the question means maximum control on the road that has the coefficient of friction of .2 or else why would mu be providedDoc Al said:
Note that when you consider friction, you have a range of acceptable speeds. Friction will give you the maximum and minimum values of the speeds that can be maintained without slipping. I would interpret the angle of maximum control to be that angle where friction is not needed.spartacus said:
Because this could be the first part of a multipart problem. Is it?spartacus said:
That is incorrect. Friction acts parallel to the surface.spartacus said:
I'm not giving up on you. :)spartacus said:
It's not a multipart problem no, how would i incorporate force of friction? would it be mg sintheta-mu?Doc Al said:
i have it right here let ill upload a screenshotDoc Al said:
No, it's an online assignmentDoc Al said:
If it's parallel should i use mu*mg sintheta then because that would make the problem a lot more difficultDoc Al said:
It seems like a poorly worded problem to me.spartacus said:
The normal force does not equal mg.spartacus said:
i mean having Ncostheta=mg for the vertical and Nsintheta-mu*mg*sintheta=(mv^2)/r for horizontalDoc Al said:
lol, i'll definitely get clarification from my teacher monday morningDoc Al said:
Two problems. Since friction has both vertical and horizontal components, your first equation would have to be modified. And, once again, the normal force does not equal mg.spartacus said:
Good idea!spartacus said:
Centripetal force on a banked road is the force that acts on a vehicle as it travels around a curved path on a banked road to keep it moving in a circular motion.
The centripetal force on a banked road can be calculated using the formula: F = (mv^2)/r, where F is the force, m is the mass of the vehicle, v is the velocity, and r is the radius of the curve on the banked road.
The angle of the banked road affects the centripetal force by providing a component of the force that acts perpendicular to the road's surface, helping to keep the vehicle in its circular path without slipping or skidding.
If the centripetal force is too low on a banked road, the vehicle will not be able to maintain its circular path and will either slide down the bank or slip off the road entirely.
The speed of the vehicle directly affects the required centripetal force on a banked road, as a higher speed will require a greater force to keep the vehicle in its circular path without slipping or skidding.