Calculating Net Work on a Moving Cargo Canister: Easy Homework Problem 1

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Homework Help Overview

The problem involves calculating the net work done on a cargo canister by three horizontal forces acting on it as it moves across a frictionless floor. The forces have specified magnitudes and angles, and the task is to determine the net work over a given displacement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of work using the formula W=Fdcostheta and question the assumptions made regarding the direction of movement. There are inquiries about the correct method for finding the net force and its components, as well as how to apply the work-energy principle correctly.

Discussion Status

The discussion is ongoing, with participants providing diagrams and exploring different approaches to calculating the net work. Some guidance has been offered regarding the need to consider the direction of forces and the resultant net force, but no consensus has been reached on the correct answer or method.

Contextual Notes

There are uncertainties regarding the angles of the forces and how they should be interpreted in the context of the problem. Participants are also navigating the implications of the canister's movement direction on their calculations.

MFlood7356
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1. Figure 7-30 shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 2.28 N, F2 = 4.54 N, and F3 = 8.24 N, and the indicated angles are θ2 = 51.6° and θ3 = 35.8°. What is the net work done on the canister by the three forces during the first 2.70 m of displacement?



2. W=Fdcostheta
Wnet=W1+W2+w3



3. I really thought this was an extremely easy problem to do but apparently not. Here's my attempt. Could someone please tell me what I did wrong?

W1=(2.28)(2.70)cos0=-6.156J
W2=(4.54)(2.70)cos51.6=-7.614J
W3=(8.24)(2.70)cos35.8=18.045J
Wnet=(18.045)+(-6.156)+(-7.614)= 4.28J
 
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Perhaps you could post or at least explain the diagram. It is a bit easier to find the resultant (net) force: \sum \vec F_i = \vec F_{net} and then use:

W = \vec F_{net} \cdot \vec d

AM
 
Last edited:
Here's a diagram: http://imgur.com/DAJ8G.gif

So would I find the magnitude of the components add them up then multiply that by the distance?
 
MFlood7356 said:
Here's a diagram: http://imgur.com/DAJ8G.gif

So would I find the magnitude of the components add them up then multiply that by the distance?
Well, you would find the magnitude of the net force by adding up the force components. Then use work = force*distance.

Your earlier attempt (shown in post #1) assumed the object moved in the +x direction, which isn't necessarily true.
 
so i would find the y and x components for each of the forces then add them up and use Pythagorean theorem then multiply that by the distance?

For the angle of F2 should it be 270-51.6=218.4 degrees and should angle of F1 be 180 degrees?
 
Yes to all statements.
 
I got 5.86 J does that sound right?
 
I don't know, I haven't worked through the problem myself and you didn't show your work. So I really can't say at this point ... but you could remedy the situation :wink:
 

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