Easy Homework Problem

  • Thread starter MFlood7356
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  • #1
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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
 

Answers and Replies

  • #2
Andrew Mason
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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: [itex]\sum \vec F_i = \vec F_{net}[/itex] and then use:

[tex]W = \vec F_{net} \cdot \vec d[/tex]

AM
 
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  • #3
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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?
 
  • #4
Redbelly98
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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.
 
  • #5
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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?
 
  • #7
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I got 5.86 J does that sound right?
 
  • #8
Redbelly98
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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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