Finding the acceleration of a crane trolley and it's load.

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SUMMARY

The discussion centers on calculating the acceleration of a crane trolley and its load, specifically an object weighing 8526 N at a 5.0° angle. The solution utilizes Newton's second law (F = ma) and the equation 9.8 / tan(90-5) to derive an acceleration of 0.857 m/s². Participants clarify that gravity does not directly affect horizontal acceleration but does influence the work done when the trolley moves vertically. Understanding the relationship between vertical and horizontal components is crucial for solving such dynamics problems.

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  • Newton's second law of motion (F = ma)
  • Basic trigonometry (specifically tangent functions)
  • Free body diagram analysis
  • Understanding of forces acting on objects in motion
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Dylan.Wallett
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Homework Statement


An object of weight 8526 N hangs at an angle of 5.0° when the crane’s trolley at point P moves to the right with constant acceleration, as shown in the diagram.
Calculate the acceleration of the trolley and load.


Homework Equations


9.8 / tan (90-5) = 0.857 m/s^2


The Attempt at a Solution


I am looking for the intuition to the problem.
I have worked out that because there is no resistance in the equation so that the weight of the load does not effect horizontal acceleration only the work done to get the trolley up to acceleration magnitude.

I would like to know intuitively why gravity effects the horizontal acceleration of the trolley. I would appreciate your help.
 
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Welcome to PF!

Hi Dylan! Welcome to PF! :smile:

As is usual in dynamics questions, you only need to use good ol' Newton's second law F = ma

draw a free body diagram and take components in a convenient direction :wink:
 
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Likes   Reactions: Irfan Nafi
gravity does not affect horizontal acceleration (except through friction, which we are ignoring) but the acceleration is not completely horizontal. since the trolley is increasing its altitude (i assume, you do not indicate whether moving to the right is going up or down) some work must be done against gravity.
 

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