Solve Statics w/Friction Homework: P Force for Cabinet Movement

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Discussion Overview

The discussion revolves around a statics homework problem involving a cabinet mounted on casters, focusing on the forces required to move the cabinet under different conditions of friction and caster locking. The participants explore the implications of static friction and the mechanics of forces acting on the cabinet.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • The cabinet weighs 120 lbs, and the coefficient of static friction is 0.30. The problem requires determining the force P needed to move the cabinet under three different scenarios regarding the locking of casters.
  • One participant describes their approach to part (b), indicating that they considered the forces acting at point B and assumed that the normal force at point A is zero due to the caster being free to rotate.
  • The participant calculated the force P by summing forces in the y-direction to find the normal force at point B and then taking moments about point A, leading to a calculated value of 45 lbs for P, which they believe should be 30 lbs.
  • Another participant inquires about the presence of a diagram to better understand the setup of the problem.
  • A participant shares a diagram and mentions trying different techniques to solve part (b), expressing uncertainty about taking moments about point A due to the line of action of the force not passing through that point.
  • One participant notes that the sum of the normal forces must exceed 120 lbs, suggesting that the front wheels may dig into the ground more as force P is applied.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the calculations and methods used, particularly in part (b). There is no consensus on the correct approach or the resulting value of force P, as different techniques yield varying results.

Contextual Notes

Participants have not resolved the assumptions regarding the line of action for the forces and the implications of the casters being locked or free to rotate. The discussion reflects different interpretations of the mechanics involved.

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Homework Statement


A 120lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h=32in, determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the caster at B are locked and the caster at A are free to rotate, (c) if the casters at A are locked and the caster at B are free to rotate.

Homework Equations


For part (a) sum of the forces in x and sum of the forces in y. I solved part (a), though I am stuck on part (b)

The Attempt at a Solution


My attempt solution for (b)
So since I know that the wheel at A is free to move then Fm2 and Fn2 are equal to zero. So the only point I am looking at is point B where there is a normal force called Fn1 and Fm1 (impending force). There is a force P (what I am looking for) acting to the right 32in from the floor. The distance between the wheels are 24in. A little later today I attach a file with my solutions for part (a) and my attempt solution for part (b). I will give you a little summary as to how I solved part (b). So I sum forces in Y to find Fn1, once that was obtained I took the moment about point A assuming that the Fm1 acting at roller B is going right through the location I took moments about, therefore it goes away? After I took moments about point A I got a result of 45lb for P. The answer should be 30lb.
 
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Is there a diagram?
 
Here is the diagram I drew. For part (b) I was just trying different techniques. If I make the small equations I still get the same results if I solve for individual forces separately. I have a feeling I can't take moments about A because the line of action for Fm1 doesn't go through the point A where I am summing moments?
 

Attachments

FN2 + FN1 > 120lbs. ... the front wheels dig into the ground more due to force P.
 

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