The problem involves a car crossing a hump-backed bridge, modeled as a circular arc with a radius of 35m. The question focuses on determining the maximum speed at which the car can traverse the bridge without losing contact at the highest point.
The discussion is ongoing, with participants exploring the relationship between the forces involved and questioning assumptions about the normal force when the car is at the point of losing contact. Some guidance has been offered regarding the nature of centripetal force and its role in the scenario.
Participants note the absence of specific values for speed, angular velocity, or mass, which raises questions about how to proceed with the analysis. The discussion also reflects on the implications of the normal force being zero at the point of losing contact.
You're supposed to find the speed!Ch3m_ said:I've tried using the equation above but no speed, angular velocity or mass is given.
Do a force analysis of the car as it tops the bridge. Hint: What force tells you when the car is just about to leave contact with the ground?Ch3m_ said:Am I missing something?
Doc Al said:Do a force analysis of the car as it tops the bridge. Hint: What force tells you when the car is just about to leave contact with the ground?
Yes, the normal force that the bridge exerts on the car. What will its value be when the car is barely making contact with the road? What will ΣF be in that case?Ch3m_ said:The reaction force..?
Doc Al said:Yes, the normal force that the bridge exerts on the car. What will its value be when the car is barely making contact with the road? What will ΣF be in that case?
You are making a false assumption. Start from scratch:Ch3m_ said:It must be greater than the weight of the car = mg? I don't know how to get an exact value though
You have the normal force acting upwards, the cars weight acting downwards and the centripetal force acting downwards?What are the forces acting on the car at the highest point?
a = v2/r ?If the car is moving at speed v, what is the acceleration of the car?
Weight + centripetal force so mg + (mv2)/r ?haruspex said:If the car went a little bit too fast it would leave the ground (you are told). What would the reaction force be then?
No, centripetal force is not an applied force. It is that component of the resultant force which produces the centripetal acceleration.Ch3m_ said:You have the normal force acting upwards, the cars weight acting downwards and the centripetal force acting downwards?
Yes, that is the normal force whilst in contact with the ground. But if it just on the point of losing contact, what extra does that tell you about the normal force?Ch3m_ said:a = v2/r ?
Weight - centripetal force so mg - (mv2)/r ?
Yes.Ch3m_ said:Wait so is it v2=gr?
Well, it is because the normal force is zero at that point.Ch3m_ said:Thanks, that must mean the normal force = 0 when the car is on the point of losing contact yeah?