Wheel overcome obstacle at speed.

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To calculate the force exerted by the obstacle on the tire, the reaction force can be determined using the wheel's mass, velocity, and the height of the obstacle. Given the parameters, the normal force can be calculated, which acts on both the vertical and horizontal axes during impact. The tire's pressure, stiffness, and suppression will influence the amount of force absorbed, but a portion will remain unabsorbed, contributing to the torque. The normal force does perform work, particularly in lifting the wheel and dissipating kinetic energy. Understanding the relationship between force and work is essential for analyzing the impact dynamics.
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Hi there, I have a project of torsion bar. I need to know value of force is making a torque on it.

So I have a velocity V=40km/h, mass per wheel m=530kg, radius of wheel R=290 mm and heigh of obstacle h = 6 cm. What will be the reaction N, on both axis, of obstacle to tire? I have no idea how to calculate that.

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Also, tire pressure is p = 0,2 MPa, tire stiffnes k = 150 000 N/m and tire supression is c= 150 Ns/m as tire will absorb some of impact force, but what value of force won;t be absorbed?
 
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Does the normal force do any work?
 
At axis Y it lifts wheel and axis X it is responsible for loosing some kinetic energy.
 
Is that a "yes"?
Do you know a relationship between force and work?
 

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