# Homework Help: Resolving components

1. Mar 6, 2013

### tooperoo

1. The problem statement, all variables and given/known data

I know this is easy, but i cant seem to sort out how to structure it properly.

This is the question:
A man is pole vaulting. Also ignore air resistance.
A man exerts a force of 810 N on a pole, at an angle of 80 degrees below the horizontal.
If the man's weight is 600 N, what is the net force acting on the man as he pushes the pole away? Include the direction and the magnitude.

2. Relevant equations

3. The attempt at a solution

Would the net force by 810 + 600? And isn't it meant to have a direction as well? Sorry, im a bit confused.

Last edited: Mar 7, 2013
2. Mar 7, 2013

ehild

3. Mar 7, 2013

### tooperoo

Im struggling to deal with the weight component.

Would you assume it has no horizontal component and only a verticle one?

4. Mar 7, 2013

### tooperoo

What I've tried:
Solved the Force (F1) with the 810N into components:

F1x = 810 sin 80 = 797.60
F1y = 810 cos 80 = 140.66

and the weight into components

Wx = 0 (is this right? there's no horizontal component because it's pointing down?)
Wy = 600N

= (797.69, 740.66)

applied Pythagoras to get the total magnitude
magnitude = sqrt( 797.69^2 + 740.66^2)
= 1088.48N

and then finding the direction

theta = tan^-1 (y/x)
theta = tan^-1 (740.66/797.69)
theta = 42.88 degrees

thus, the total force acting on the man is 1088.48N in the direction of 42.88 degrees below the horizontal

Last edited: Mar 7, 2013
5. Mar 7, 2013

### ap123

What direction are your x- and y-axes pointing in?

The problem says that the weight is 500N

6. Mar 7, 2013

### tooperoo

bugger. in my horrible handwriting, ive made a 0 a 6. and that's a typo. ill edit it now. meant to be 600

and the usual x and y axis

7. Mar 7, 2013

### tooperoo

edited in the new corrections

8. Mar 7, 2013

### ap123

Be careful with the signs - the weight is in the negative y-direction.
Also, I think the sin and cos for the components of F1 are the wrong way round?
What angle is the force of the pole on the man making with the x-axis?

9. Mar 7, 2013

### tooperoo

ok, so the F2 should be (0, -600)?

and i thought that sin was the opposite side of the triangle to the angle you are looking at. so the opposite side to the 80 degree angle

10. Mar 7, 2013

### ap123

Yes.

F1 makes an angle of 80° with the positive x-axis.
Is this the way you've got it?

11. Mar 7, 2013

### tooperoo

yeah, it makes an angle of 80 degrees BELOW the positive x axis

and F2(0, -600) is right?

thank you so much for your help mate.

12. Mar 7, 2013

### ap123

The force of the man on the pole is 80° below the x-axis.
So, the force of the pole on the man is 80° above the x-axis.

Yes

13. Mar 7, 2013

### tooperoo

Ohhh i see. So the force is acting basically 180 degrees to what i thought.

It's the opposite reaction no?

So how does this change my equations?

(if i ever meet you in real life, you're getting a foot rub)

14. Mar 7, 2013

### ap123

Yes, this is the Newton3 reaction force - you want the force on the man.
If you draw the free-body diagram for the man you'll see your equations are almost right, but the sine and cosine are the wrong way round.

15. Mar 7, 2013

the sine and the cos are the wrong way

Last edited: Mar 7, 2013
16. Mar 7, 2013

### tooperoo

thank you so much man. i appreciate it so much