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

In summary, the conversation discussed the correct method of indicating direction using the [N 20 W] notation and clarified the formula for determining the displacement of a log being pulled by two tugboats with given forces and initial conditions. It was determined that the displacement would be [W 10 N] and a link was provided for further clarification on direction notation.
  • #1
paolostinz
25
1

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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  • #2
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
 
  • #3
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] ?
 
  • #4
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
 
  • #5
Sorry lol, let me clarify, they are meant to indicate direction; [north 10 degrees west] in reference to the log's original position.
 
  • #6
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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  • #7
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?
 
  • #8
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
 
  • #9
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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!
 

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

1. What is net force in two dimensions?

Net force in two dimensions is the sum of all the forces acting on an object in both the x and y directions. It takes into account both the magnitude and direction of each force and determines the overall acceleration of the object.

2. How is net force calculated in two dimensions?

To calculate net force in two dimensions, you must first identify all the forces acting on the object and their respective magnitudes and directions. Then, you can use vector addition to add the forces in each direction separately and determine the overall net force.

3. What is the difference between net force and individual forces?

Individual forces act on an object in a specific direction and can either increase or decrease the object's motion. Net force takes into account all the forces acting on an object and determines the overall acceleration and direction of the object's movement.

4. How does net force affect an object's motion?

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that a greater net force will result in a greater acceleration of the object.

5. Can net force be negative in two dimensions?

Yes, net force can be negative in two dimensions. This means that the overall force acting on the object is in the opposite direction of its motion. It can also indicate that the object is slowing down or changing direction.

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