# Newton's laws : some math work

1. Jun 2, 2010

### ramenluver50

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

1)The particle shown below is at rest, where F = 65.0 N, and θ = 33.1°. Find the magnitudes of F1 and F2.

2)Just after opening a parachute of negligible mass, a parachutist of mass 99.5 kg experiences an instantaneous upward acceleration of 1.09 m/s2. Find the force of the air on the parachute.
-specifically magnitude

3)Find the tension in each string in the figure below, where θ1 = 41.0°, θ2 = 20.5°, and w = 28.0 N. for t1, t2, and t3

2. Relevant equations

F=ma

3. The attempt at a solution

i dont know...

2. Jun 2, 2010

### rock.freak667

For the first one, if the particle is at rest, then what should the sum of the forces be in the x and y directions?

3. Jun 2, 2010

### ramenluver50

for the X and y for just F, i used sin 33.1=X/65 and for Y used cosine, but thats about the farthest ive went.

4. Jun 2, 2010

### rock.freak667

right, so if the particle is at rest, meaning it is in equilibrium, what should the resultant force be horizontally and vertically?

5. Jun 2, 2010

### ramenluver50

i think horizontal, but im not sure

6. Jun 2, 2010

### rock.freak667

Are you familiar with the term equilibrium?

7. Jun 2, 2010

### ramenluver50

yes, where there is balance

8. Jun 2, 2010

### rock.freak667

Right! So considering the x-direction, if there is balance, then shouldn't the forces point to the left be equal to the forces pointing to the right?

Similar for the y-direction.

9. Jun 2, 2010

### ramenluver50

so your saying if i just find the X and Y for F they X and Y should be the same force equivalence to F1 and F2? , i tried for X where sin 33.1=X/65, and X would be 35.5, apparently the answer is incorrect.
EDIT:*******
sorry, you are right, i entered the answer in the wrong answer slot, haha.. thx

10. Jun 2, 2010

### rock.freak667

Now try the second and third problems.

11. Jun 2, 2010

### ramenluver50

for Number 2,
i used the F=ma formula, where, 99.5 (1.09) to get the newtons of 108, which is wrong...

for #3, i got T3 which is 28 N, but for T1 and T2, do i just used T3 as the vertical measurement to find T1 and T2 ?

12. Jun 2, 2010

### rock.freak667

108 is the resultant force on the body, what other forces are acting on the body?

Yes, that should work.

13. Jun 2, 2010

### ramenluver50

2) there is also gravity,not sure what to do with it though,

3) i did sin 41=28/T1 which gave me 42.6 , that isnt right bc its force is larger... :-/

14. Jun 2, 2010

### rock.freak667

There is the force of gravity and what other force? (they asked you to find this force)

You have T1 and T2 acting at their respective angles. These two force both have components in the x and y directions. What are these components? If the figure is in equilibrium, the sum of forces in any direction is ?

15. Jun 2, 2010

### ramenluver50

thx i got number 3 down,

the other resistance for number 2 is air....

16. Jun 2, 2010

### rock.freak667

Good, so what directions do the weight and the force of air act?

17. Jun 2, 2010

### ramenluver50

wieght is downward force, and air is up...

18. Jun 2, 2010

### rock.freak667

Then what would be an expression in terms of these two for the resultant force?

19. Jun 2, 2010

### ramenluver50

Newtons...?

20. Jun 2, 2010

### vorcil

A force is a vector quantity,

which means it has a magnitude and a direction,

in your first question you are given the magnitude of the force = 65 N and the direction to be 65degrees
in the x&y direction

you can separate this force into the two forces that make that force i.e
$$Ftotal = Fxdirection + Fydirection$$

but you're probably wondering how you get the force in the X direction from the total force?

use trigonometry , that is your F total to be the hypotenuse, Fx to be your adjacent, and Fy to be your opposite

just use basic trigonometry to figure fx and fy out

and you know that the particle is at rest,
so f1 must = -fx
and f2 = -fy
otherwise the particle would start moving!