Wednesday

July 30, 2014

July 30, 2014

Posted by **Jane** on Sunday, September 12, 2010 at 10:30pm.

i'm not sure how to start this problem

help pleasee

- physics -
**got the answer from someone**, Saturday, July 13, 2013 at 9:36pmFigure out the East-West forces and add them up:

... from F1: 40 cos(60 deg) = 20

... from F2: -90

... from F3: F3 cos(theta)

Add them up:

... -70 + F3 cos(theta)

Set it equal to zero (the ball does not move)

... 70 = F3 cos(theta)

Now do the same for the North-South forces. You should get

... 34.641 = F3 sin(theta)

We have two equations in two unknowns. We can solve them, first for the angle:

34.641 / 70 = [-F3 sin(theta) ] / [F3 cos(theta) ]

0.4949 = -tan(theta)

Using arctangent, we find

theta = 26.33 degrees.

And cos(theta) = 0.896

and sin(theta) = 0.444

Putting these back into the first two equations:

70 = F3 cos(theta)

70 / 0.896 = F3 = 78.102

And also

34.641 = F3 sin(theta)

34.641 / 0.444 = F3 = 78.102

- physics -
**got the answer from someone**, Saturday, July 13, 2013 at 9:36pmFigure out the East-West forces and add them up:

... from F1: 40 cos(60 deg) = 20

... from F2: -90

... from F3: F3 cos(theta)

Add them up:

... -70 + F3 cos(theta)

Set it equal to zero (the ball does not move)

... 70 = F3 cos(theta)

Now do the same for the North-South forces. You should get

... 34.641 = F3 sin(theta)

We have two equations in two unknowns. We can solve them, first for the angle:

34.641 / 70 = [-F3 sin(theta) ] / [F3 cos(theta) ]

0.4949 = -tan(theta)

Using arctangent, we find

theta = 26.33 degrees.

And cos(theta) = 0.896

and sin(theta) = 0.444

Putting these back into the first two equations:

70 = F3 cos(theta)

70 / 0.896 = F3 = 78.102

And also

34.641 = F3 sin(theta)

34.641 / 0.444 = F3 = 78.102

- physics -
**got the answer from someone**, Saturday, July 13, 2013 at 9:37pmthe answer above is for another calculation so plug in you values and derive your answer

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