Your link is to the hard disk of your computer. No one else has access to it.
If you wish, you could upload it to one of the many image servers available to the public free of charge, and we'd be able to see the picture through your link.
To be honest, you are generally not able to post a link. If that's the case, please post a detailed description.
could someone tell me how to do this?
1. A diagram illustrating several tugboats applying forces to a barge is shown below. What is the magnitude and direction of force does tugboat C (Fc) need to apply to keep the tugboat from moving?
C:\Documents and Settings\26536\My Documents\Downloads\sO6UEzpWM5rX2LXyP_t29Ug.png
that is the link to the picture
5 answers
oh... i feel stupid
it wont let me post the whole thing, but here is some.
ok, the picture has a rectangle in the middle, with the long sides horizontal. coming out of the left side is a "tugboat" connected to a rope. it is labelled Fa=7500 N.
ok, the picture has a rectangle in the middle, with the long sides horizontal. coming out of the left side is a "tugboat" connected to a rope. it is labelled Fa=7500 N.
here is the rest
coming out of the upper right corner of the rectangle is another tugboat. the angle between the boat and the side of the rectangle is 140 degrees. this boat is labelled Fb=5000N
Fc is sticking out of the bottom of the boat
coming out of the upper right corner of the rectangle is another tugboat. the angle between the boat and the side of the rectangle is 140 degrees. this boat is labelled Fb=5000N
Fc is sticking out of the bottom of the boat
Would it be:
"Fc is sticking out of the bottom of the barge"?
So there is a system of three forces that we would like to stay in equilibrium.
The two methods in use would be
1. the triangle of forces, by which the force vectors are joined head-to-tail. Since the direction of all three are known, as well as two of the magnitudes, the third can be found by trigonometry.
2. The other method is by resolution into the coordinate axes, namely
ΣFx=0, and
ΣFy=0.
Try one or both of the methods and post your answer for checking.
In this case, if I understand the diagram correctly, Fc is vertically "downwards", you could use the second equation of method 2 to get:
Fc*sin(90)=Fb*sin(140)
from which you can solve for Fc explicitly.
"Fc is sticking out of the bottom of the barge"?
So there is a system of three forces that we would like to stay in equilibrium.
The two methods in use would be
1. the triangle of forces, by which the force vectors are joined head-to-tail. Since the direction of all three are known, as well as two of the magnitudes, the third can be found by trigonometry.
2. The other method is by resolution into the coordinate axes, namely
ΣFx=0, and
ΣFy=0.
Try one or both of the methods and post your answer for checking.
In this case, if I understand the diagram correctly, Fc is vertically "downwards", you could use the second equation of method 2 to get:
Fc*sin(90)=Fb*sin(140)
from which you can solve for Fc explicitly.
1 answer