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physea
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What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
The friction will be directly opposite mgsinθ if the car is going slow enough, but it will be the other way if the object is going fast enough. Therefore, the equation depends on which way the car is slipping. If it is in perfect equilibrium, then mgsinθ and friction cancel out and you're left with only components perpendicular to the road (such as the normal force and mgcosθ) causing the centripetal acceleration.physea said:What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
If you want an answer to that then it is probably better if you actually define all the variables in those equations, along with a suitable diagram. Also, you will get a better response if you can manage to be a bit more polite with your replies; the customer is not always right here because there are no (paying) customers. "That's all".physea said:What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
Is the angle measured from the horizontal or from the vertical? Is positive for R1, R2 and the centripetal acceleration taken in the sense of an inward net force? I assume yes. Is positive for mg cos(θ) taken in terms of down=positive or up=positive. That would obviously have a key influence on the answer you seek.physea said:R1 and R2 are the normal reactions of the ground to the vehicle's wheels
mg is the vehicle weight
V it's speed
θ the angle of the banked road
The normal reaction in a banked road circular motion problem refers to the force exerted by the road surface on a vehicle as it moves in a circular path. This force is perpendicular to the surface of the road and is responsible for keeping the vehicle from sliding off the road.
The angle of banking has a significant impact on the magnitude of the normal reaction in a banked road circular motion problem. A steeper banked angle will result in a higher normal reaction force, while a shallower angle will result in a lower normal reaction force.
Friction plays a crucial role in the normal reaction in a banked road circular motion problem. It acts in the direction opposite to the motion of the vehicle and helps to balance the forces acting on the vehicle, thereby keeping it on the road.
The normal reaction will change as the speed of the vehicle changes in a banked road circular motion problem. As the speed increases, the normal reaction force decreases, and as the speed decreases, the normal reaction force increases. This is due to the changes in centripetal force and centrifugal force as the vehicle accelerates or decelerates.
Yes, the normal reaction force can be greater than the weight of the vehicle in a banked road circular motion problem. This is possible when the banked angle is steep enough and the speed of the vehicle is high enough to generate a large enough centripetal force to counteract the weight of the vehicle.