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Force, friction?

Force, friction? Topic: Problem solving wheel what can i do
June 17, 2019 / By Hanniel
Question: If someone can write these out and the solution so I can solve the similar problems it would be great, thx! Suppose a 40kg cart is sitting on a loading ramp with a 15* incline. What’s the ramp’s normal force? Suppose the cart described above has rubber wheels. Whats the static and kinetic friction between the wheels and the concrete ramp after a good rainstorm?
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Best Answers: Force, friction?

Edwin Edwin | 8 days ago
Draw a FBD of the cart on the inclined ramp. The y axis and x axis will be tilted a bit (x-axis is along the ramp) Forces in the y direction will be zero. Thus do a Fnet = 0 for just the vertical right now. There are 2 forces here, vertical component of gravity and the normal force. Fnety = ma N - mgcosθ = 0 N = mgcosθ N = (40kg)(9.8)cos15 And that's your normal force on the ramp. For the next part, you'll be looking at the x-axis now solely. To find the static friction, again set Fnet = ma to zero. This will get you the maximum amount of static force the ramp can support. 2 forces here in the x direction, friction and the horizontal component of gravity. mgsinθ - friction = 0 mgsinθ = friction I guess you can end it there, but you can further simplify if you wanted by doing this: Let friction = Us*N where N is the normal force mgsinθ = Us*mgcosθ mass and gravity cancel out and sin / cos gives you tanθ tanθ = Us tan15 = Us So that's the maximum coefficient of static friction that was found. Don't think you have enough information to get the kinetic friction, but do realize that once you go 1 iota past that static friction coefficient, then you will have movement along the incline and thus you no longer are working with static friction, but with kinetic friction.
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We found more questions related to the topic: Problem solving wheel what can i do


Edwin Originally Answered: PHYSICS HOMEWORK FORCE OF FRICTION HELP?
Here, a=2m/s^2 F=45N net force on the wagon=ma =15*2=30N now,45-f=30 or,f=15N
Edwin Originally Answered: PHYSICS HOMEWORK FORCE OF FRICTION HELP?
F_x - F_friction = m * a. so 45 - F_friction = m * a = 15kg * 2 m/s^2 F_friction = 45N - 30N = 15N I majored in physics, but found a nice tutorial on how to better understand, and set up these types of problems. Check out this link: http://www.physicstutorials.org/home/mechanics/dynamics/friction-force
Edwin Originally Answered: PHYSICS HOMEWORK FORCE OF FRICTION HELP?
A skydiver falls at a relentless velocity after his/her parachute opens. If their finished mass (alongside with chute) is ninety kg, what's the stress of drag on him/her? _90 x 9.80 one = 882.9 N..................ans coefficient of friction = Frictional stress / well-known reaction ?s = 12 / ( 6 * 9.80 one) = 0.204..................ans ?k = 10 / (6 * 9.80 one) = 0.17 .....................ans

Edwin Originally Answered: Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum?
Fnet = ma The only force acting on the car is the force of friction. The magnitude of this force is equal to: Ff = uFn, where Fn is the normal force On a flat surface, the normal force must balance the car's weight since it does not accelerate up or down. Therefore, Ff = umg An object moving in a circle at constant speed will have acceleration: a = v²/r, where v is the object's speed and r is the radius of the circle If Fnet = ma, then: Ff = mv²/r umg = mv²/r ug = v²/r The masses cancel out, which makes sense because you cannot combine [metre] and [kg] to get [metre/second]. ugr = v² v = sqrt(ugr) The rest is a no-brainer. If you can solve a problem using variables, then you can solve the problem given any set of numbers. The point is to simplyfing a hard problem (not straightforward) to a problem that is straightforward, where you can substitute numbers into a formula and evaluate. This makes your work much easier to read and it makes it easier to check for errors. A useful technique to check whether or not your answer makes sense is dimensional analysis: basically if the units in your answer don't match the units of the required quantity, then you know you've done something wrong. Let's anaylse the answer to this problem: units for speed: m/s units for: sqrt(ugr) sqrt(m/s² m) - the coefficient has no units = sqrt(m²/s²) = m/s The units match, so the answer makes sense.
Edwin Originally Answered: Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum?
The maximum amount of force that friction can provide is: F_max = (normal force)×μ = mgμ The force required to go in a circle is: F_required = ma = m(v²/r) F_max has to be greater than or equal to F_required: F_max ≥ F_required mgμ ≥ mv²/r Solve the inequality for "v". You are given "r" and "μ", so plug those in.

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