Distance traveled by a accelerating truck with steel load.

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SUMMARY

The discussion focuses on calculating the minimum stopping distance for a truck carrying a 10,000kg steel load, which is not secured and has a coefficient of static friction of 0.500 with the truck bed. The truck, weighing 20,000kg, is initially moving at 12.0m/s. The solution reveals that the minimum stopping distance to prevent the load from sliding forward is 14.7 meters. Key equations utilized include the static friction formula and kinematic equations to derive the necessary acceleration and stopping distance.

PREREQUISITES
  • Understanding of Newton's 2nd Law of Motion
  • Knowledge of static friction and its coefficient
  • Familiarity with kinematic equations for motion
  • Basic principles of dynamics and forces
NEXT STEPS
  • Calculate the acceleration of the truck using Newton's 2nd Law
  • Explore kinematic equations for different motion scenarios
  • Study the effects of load securing methods on vehicle safety
  • Investigate the implications of friction coefficients in various materials
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and safety professionals focused on load management and vehicle dynamics.

blackkeys
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Distance traveled by an accelerating truck with a steel load. With a twist!

Homework Statement



Consider a large truck carrying a heavy load such as steel beams. A significant hazard for the driver is that the load may slide forward crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10,000kg load sits on the flat bed of a 20,000kg truck moving at 12.0m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.

Homework Equations



Fsf is less then or equal to the coefficient of static friction * (mass * gravity)

A= (velocity final - velocity initial) / (total time)

change in x = original velocity in the x direction + acceleration in the x direction * time


The Attempt at a Solution



I know that both masses of the beams and truck are not relevant to the solution to the problem. However that's all I've been able to figure out. I think that if the truck is moving at 12m/s the bars must be resisting motion at 12m/s. Is this true? I wasn't sure which direction to go after determining that mass wasn't going to be any help. How could I use this to help solve the problem?

The solution to the problem is 14.7m
 
Last edited:
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blackkeys said:

Homework Statement



Consider a large truck carrying a heavy load such as steel beams. A significant hazard for the driver is that the load may slide forward crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10,000kg load sits on the flat bed of a 20,000kg truck moving at 12.0m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.

Homework Equations



Fsf is less then or equal to the coefficient of static friction * (mass * gravity)

A= (velocity final - velocity initial) / (total time)

change in x = original velocity in the x direction + acceleration in the x direction * time


The Attempt at a Solution



I know that both masses of the beams and truck are not relevant to the solution to the problem. However that's all I've been able to figure out. I think that if the truck is moving at 12m/s the bars must be resisting motion at 12m/s. Is this true? I wasn't sure which direction to go after determining that mass wasn't going to be any help. How could I use this to help solve the problem?

The solution to the problem is 14.7m
The acceleration of the beam load must be the same as the truck's acceleration in order for the load not to slide with respect to the truck. You'll need to calculate that acceleration of the load using Newton's 2nd law. The use your kinematic equations...you made in error in one of them for delta x.
 

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