If a car is moving with constant speed in a circle. What force draws the car inward? Explain with a drawing.
Let's put all the mathematical calculation away and discuss about the idea of " Centripetal force ". In physics , we use the term " Centripetal force " to describe any forces in the Universe that help something in performing a circular motion. A centripetal forces could be a reaction force , gravitational force or maybe even frictional force. Whatever forces. Any forces that help an object to move in circular path is called the centripetal force. Now let's look into a simple situation.It's a rubber stopper swirling around the stick , bonded by a cord. It's obviously going in a circular motion. There must be a centripetal force acting on the rubber stopper , otherwise it wouldn't go in circular path. In this case , the force that helps the rubber stopper to remain travelling in a circular path is the tension of the cord. Without the cord , the rubber stopper would fly away you can say. Hence , the centripetal force of this case is the tension that pulling the rubber stopper inward. Now let's see another one. Owh , a car going around like that. How ? Yes of course , good driving skill plus the frictional force that prevents the car to skid away. Just imagine what happen if the tyres of the car are smooth? I am sorry , but you won't see the car do this due to the absent of centripetal force ( frictional forces of those tyres ). One more if you don't mind. Okay , I like this. A satellite orbiting the Earth. Can you imagine what will happen if the gravity of the Earth suddenly disappears ? ( Ignore what would happen to the Earth and your wife in it.) The satellite won't orbit us anymore. Maybe it goes straight to the Sun , I don't know. The centripetal force ( gravitational force ) however helps in making the satellite to orbit.
Describe the forces involved when a car suddenly stops on a wet road.?
for a car, you have various forces... forward momentum / inertia, air resistance, gravity, etc... when a car is moving, there is inertia and forward motion... friction from the wheels move the car along the road, but when the road is wet, and you force the brakes, the tires lose traction , aka, the friction between the tire surface and the road surface due to viscosity (i.e. the road is lubricated by the water) once this happens you slide.... it is also possible to hydroplane. basically what that is caused by, is when water forces itself between the tire and the road surface causing you to slide on the water itself and have no contact with the road surface. you're basically driving on liquid ice for that moment, or, more accurately, you're tires are doing the exact same thing as when you get on a surf board, or water ski's. that's the dumbed down explanation. it won't get you an A on your test, but, it's an accurate explanation
How do I calculate the average braking force on a car of mass 800 kg initially moving at 8.9 m/s?
T To answer this question we require the time .Lets assume its one second.Mass = 800 kgInitial velocity = 8.9 m/sFinal velocity = 0 m/sTime = 1 sBRAKING FORCE = Mass × AccelarationAcc. = v-u ÷t= -8.9Thus , force = 7120
A 1500 kg car skids to a halt on a wet road where friction coefficient = 0.48.?
This problem can successfully be solved using Newton's First Law --> Fnet = ma m=1500 kg, coefficient of friction = .48, Vfinal = 0 m/s, and assuming distance traveled = 69 m. Fnet = (m)(a) (u)(m)(g)=(m)(a) , m can be cancelled. a = (.48)(9.8) = 4.704 m/s^2 Vfinal^2 = Vinitial^2 + 2ad 0 = (v)^2 + 2(-4.704)(69) -(v)^2 = -(2)(4.704)(69) , minus signs cancel Vinitial = 25.5 m/s
A 1000 kg car traveling at a speed of 22 m/s skids to a hault on wet concrete.?
Halt (which does not mean stop!). (1000 x 9.8) x 0.2 = 1,960N. braking force. Accelertaion = (f/m), = 1960/1,000, = 1.96m/sec^2. Distance to stop = (v^2/2a), = 123.47 metres.
Please help with a SIMPLE physics question involving friction!?
A car is traveling at 52.0 mi/h (23.25 m/s) on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.098, what is the minimum distance in which the car will stop? ( in m) (b) What is the stopping distance when the surface is dry and µs = 0.600? (in m) I've been working on this problem for a long time and can't seem to get anywhere without knowing the mass of the car.. which isn't given... can someone please help me? Any help is greatly appreciated! I know f=uN and that you can do sum of forces in x direction and then sum of forces in y direction but I can't figure out how to combine these equations to get the answer...
In a moving car the wheel will skid if the brakes are applied too suddenly. What is the cause of this?
Basic physics. The maximum force the wheels can exert on the car to stop it is the weight of the car multiplied by the coefficient of friction between the tires and the road surface. That depends on the tires, the road, and whether it’s wet or icy, but it’s around 1.0. The maximum force that a brake can exert on a wheel to stop it rotating depends on the coefficient of friction between the brake pads and brake rotor. Let’s say around 1.0 too. But it depends not on the weight of the brakes, but only on how well they are made - the brakes push against a C-shaped metal shell; it’s like cracking a nut in your teeth instead of standing on it. The braking force can easily exceed the tire friction - you want a small driver to be able to stop a heavy car going downhill on a dry road, as well as a strong driver able to stop a car in the wet. So when the brakes are applied strongly, the wheels will skid. Sliding friction is typically a bit less than static friction, so once skidding, the wheel will continue to skid until the brake pressure is reduced, either by the driver or by ABS.
You are driving a 2500.0 kg car at a constant speed of 14.0 m/s along a wet, but straight, level road.?
As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0 m. What is the coefficient of kinetic friction between your tires and the wet road?