What is the Work Involved in Lifting and Moving a Box with a Crane?

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    Crane Kinematics
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Homework Help Overview

The discussion revolves around calculating the work involved in lifting and moving a box with a crane, specifically focusing on the forces at play and the work done in both vertical and horizontal movements. The subject area includes concepts from mechanics, particularly work and force.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to break down the work into components related to lifting and moving the box, questioning how to find the necessary forces and the acceleration involved. Some participants suggest considering the force required to lift the box and clarify the relationship between the forces acting on it.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved in lifting the box. Some guidance has been offered regarding the assumptions about the lifting force and its relationship to the weight of the box, but no consensus has been reached yet.

Contextual Notes

Participants are working under the assumption that there is no friction involved in the horizontal movement and are discussing the implications of lifting the box slowly. The original poster expresses uncertainty about the acceleration and its role in the calculations.

Adrian379
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1. The problem statement, all variables and given/known
G=8400N
h=35m
d=10m
g=10m/s^2

A crane elevate a box with G=8400N at h=35. After this, the box is moved horizontal on distance 10m. No friction.

Work = ?
ps. Sorry for bad English.

Homework Equations


I think that W=W1 ( work on elevating ) + W2 ( work on moving ).
W1=F1*h and W2=F2*d

In the first I must tu find F1.
F=F1-G ==> F1=G-F=G-ma

Now, F2:
F=F2==> F2=m*a.

W=F1+F2=G-ma+m*a.

But I don't know a...

The Attempt at a Solution


[/B]
I know I am bad but I want to learn. What can I do to solve this exercice ? Thank you very much !
 
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Hello Adrian, :welcome:

Assume the lifting goes very slowly. So the only force that is needed for that is ##mg## = G . Keep the direction of the force in mind.
(why do I say that?).
 
Because F = -G ? Or.. F is a little bit stronger that G to lift ?

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Last edited:
//Sorry for x2 post.
 
Adrian379 said:
Because F = -G ?
That is what is meant, yes.
 

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