# Tension and force

Hi I have an Exam on Friday and really could use help with two area's. Thay are Force and Tension.

The problem I have with force is when you have more than 2 force's. This is one example that I have in my book. "A horse is stuck in the mudd with a F1=1500N [w20*]
F2=2000N [e30*]
F3=300N
I can't do this problem at all. The answer is writen down rong so its of no help Could someone explain how to do this problem?

The problem I have with tension is that I dont understand it at all. I need to know how to find tension from E to W (vice verca) and N to S. I also need to know how to find tension for an example like this.

M1=35kg
M2=20kg
The man is pulling with a 200N force at a 40* angle.

Thanks a lot for any help, I really need it or I might fail

Can anyone help me with this?

For more than two forces. Write the forces in the satndard vector notation and add them to get the resultant

$$\sum_{i=1}^{n} \vec F_i= \vec F_1+\vec F_2 + ....... + \vec F_n$$

Last edited:
HallsofIvy
Homework Helper
People might be able to help more if you wrote the problems more clearly. For example, in the first one, you never said clearly what it was you were asked to do! I can guess that you are asked to find the net force on the horse but it would have been nice if you had told us that. Also, I can guess that by "F1=1500N [w20*]" you mean a force of 1500 Newtons, at an angle of 20 degrees west of north (I choose north just because it strikes me as a reasonable base) but I'm not at all certain of that!

Assuming that the forces are: F1= 1500 N at 20 degrees west of north, F2= 2000 N at 30 degrees east of north, and F3= 300 N directed south, there are two ways to find the "resultant" (the net force).
One is to draw a picture, with the force "arrows" connected, break it into triangles, and use trigonometry to find the length and angle of the resulting sum of vectors.
Simpler for a more complicated problem like this is to break each vector into components and add the components.
F1 has "length" 1500 and makes an angle of 20 to the west of north. I set up my coordinate system with the positive x-axis east and positive y-axis north (that's the standard convention- other choices would give the same answer). Now I have a right triangle with hypotenuse 1500 and angle of 20 degrees. The component pointing due north is the "near side" and cosine= "near/hypotenuse" so "north"/1500= cos(20)= 0.939. "north"= 1500(0.939)= 1409.
The component pointing due west is the "opposite side" and sine= "opposite/hypotenuse" so "west"/1500= sin(20)= 0.342 or "west"= 1500(0.342)= 513. Since I chose east as the positive x-axis and this is pointing west, it is actually -513. With this coordinate system,
F1= (-513N,1409N).
The second force, F2, has "length" 2000 and makes an angle of 30 degrees. Using the same reasoning, "north"/2000= cos(30)= 0.866 so
"north"= 2000(0.866)= 1732 and "east"/2000= sin(30)= 0.50 so "east" = 1000. F2=(1000N,1732N).
Finally, the third force, F3, has "length" 300 and points due south. That's easy. F3= (0, -300).

The resultant force is (-513+ 1000+ 0,1409+ 1732- 300)= (487, 2841).

Now, we ought to put that answer back into the same "terms" as the forces were given. Drawing a line 487 "long" to the east and 2841 "long" to the north, we have a right triangle with legs of length 487 and 2841 and so (by the Pythagorean triangle), hypotenuse of length &radic;(4872+ 28412)= 2822.
Since tangent= "opposite/near", the tangent of the angle the line makes with north has the east, x, component as "opposite" and the north, y, component as "near" so the tangent of the angle is 487/2841= 0.171 and, taking the arctangent of that, the angle is 9.7 (i.e. approx 10 degrees). The resultant vector has magnitude 2822 N and makes an angle of (approximately) 10 degrees to the east of north.

I have absolutely no idea what the second problem is supposed to mean. You give two masses and then say "The man is pulling with a 200N force at a 40* angle." Pulling what? What do the two masses have to do with the man pulling?? COPY THE PROBLEM EXACTLY AS IT IS GIVEN!

The easiest way to do tension and force problems is to draw SCALED diagrams. Place your forces tip to tail. From your starting point to the end of your last arrow is your resultant. Using Trig and the laws of sins/cosines you should be able to solve for the angles. If you are asked to find the equilibrant you simply take the force and direction from your resultant and add 180 degrees. By adding 180 degrees that will make the force point in the opposite direction therefore putting the system in equilibrium. Hope this helps.