What is the net force acting on the log after 10 seconds?

AI Thread Summary
The discussion centers on calculating the net force acting on a log being pulled by two tugboats, with a mass of 250 kg. The calculated resultant force is 1164 N, but there is confusion regarding the correct angle for its direction. Participants clarify that the angle should be expressed as [N 10 W] or [W 10 N], depending on regional conventions for indicating direction. The focus shifts to determining the log's displacement after 10 seconds, emphasizing the need to calculate acceleration from the net force. Ultimately, the conversation highlights the importance of consistent directional notation in physics problems.
paolostinz
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Homework Statement



Two tugboats are pulling a large log, as shown in the following diagram. The log has a mass of 250 kg and is initially at rest. How far have the tugboats moved the log after 10 s?

http://imgur.com/GS7Y80x

Homework Equations



c^2=a^2 + b^2 -2ab cosC

sin A/ a= sin B/ b= sin C/ c

Fg=G m1 m2/d^2

F=ma

d=1/2 a t^2



The Attempt at a Solution



c^2=(400 N)^2 + (800 N)^2 -2(400 N)(800N)cos150
c=1164 N

sin A/ 400 N= sin150/ 1164 N
A=sin^-1 (400N sin150/1164N)
A=10 degrees

Here's where I'm having the most trouble, in my book it shows that you subtract the 20 degrees from angle A (10 degree angle). I don't understand this, how does it reflect the angle to which the net force is applied to?

So my answer for the first part is 1164 N [ N 10 W].
 
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paolostinz said:

Homework Statement



Two tugboats are pulling a large log, as shown in the following diagram. The log has a mass of 250 kg and is initially at rest. How far have the tugboats moved the log after 10 s?

http://imgur.com/GS7Y80x

Homework Equations



c^2=a^2 + b^2 -2ab cosC

sin A/ a= sin B/ b= sin C/ c

Fg=G m1 m2/d^2

F=ma

d=1/2 a t^2



The Attempt at a Solution



c^2=(400 N)^2 + (800 N)^2 -2(400 N)(800N)cos150
c=1164 N

sin A/ 400 N= sin150/ 1164 N
A=sin^-1 (400N sin150/1164N)
A=10 degrees

Here's where I'm having the most trouble, in my book it shows that you subtract the 20 degrees from angle A (10 degree angle). I don't understand this, how does it reflect the angle to which the net force is applied to?

So my answer for the first part is 1164 N [ N 10 W].

A is the angle between the 800 N force and the resultant force. The 800 N force is at 20°angle to the horizontal. So the angle of the resultant force with respect to the horizontal is 20° - A.


ehild
 
So what you're saying is I determined the wrong angle?
Or did I even need to determine the angle?

Should my answer be 1164 N [N 20 W] ?
 
paolostinz said:
So what you're saying is I determined the wrong angle?
Or did I even need to determine the angle?

Should my answer be 1164 N [N 20 W] ?

You determined the angle between the resultant and the 800 N force, and got A=10°. You need to give the direction of the resultant force. Which is (20-A), that is 10° to North with respect to West (I am not sure what [N 20 W] means :redface:. 20-A happens to be the same as A: 10°.


ehild
 
Sorry lol, let me clarify, they are meant to indicate direction; [north 10 degrees west] in reference to the log's original position.
 
Well, we do not indicate directions that way in my country (Hungary) . So does [N 20 W] mean the direction of the red arrow in my picture? If so, your answer is correct.


ehild
 

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Yes, that is what I meant. I'm not sure if that's the correct way to reference it, it's just the only way I thought of describing it. Thank you so much for your help!

Btw, how do you attach thumbnail like that?
 
paolostinz said:
Yes, that is what I meant. I'm not sure if that's the correct way to reference it, it's just the only way I thought of describing it. Thank you so much for your help!

Btw, how do you attach thumbnail like that?

I draw the picture in Paint, save as jpeg on my computer, then Go to Advanced, scroll down to Attach Files, click on Manage Attachments, browse my files and then upload the picture.

I just have noticed that the question was not the force but the displacement in 10 seconds. You know the resultant force, you know the mass of the log, what is the acceleration? What is the displacement?

ehild
 
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  • #11
In the link you provided it says that north or south should be stated first always. Could you elaborate a little bit because this part has confused me in the past? I usually just go with my gut which isn't entirely accurate.
 
  • #12
Then you have to change the angle so it is measured from the North.
It will be [N 80 W].
They use this notation in the high-school physics around here. But I don't think they use the "North first" rule. But you put first the direction you measure from.
But you are right, it seems that North (or South) first may be more common.
http://www.firefightermath.org/index.php?option=com_content&view=article&id=51&Itemid=131
And the angle less than 90 degrees. At least for firefighters.:smile:

In the end is just a convention and is good to use the convention in your textbook.
 
  • #13
ehild said:
Thank you nasu, for the link ! I could not find a place explaining how to indicate direction this way.

ehild

No problem. Glad to be of (some) help.
 
  • #14
You both are/were a tremendous help, thank you!
 
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