# Mechanics / physics

• December 29th 2008, 09:48 AM
djmccabie
Mechanics / physics
The diagram shows four coplanar forces of magnitude SN, RN, 60N and 80N acting at a point 0 in the direction shown.

http://http://www.mathhelpforum.com/...1&d=1230572234http://www.mathhelpforum.com/math-he...1&d=1230572234

Given the forces are in equilibrium, find the values of R and S.

I try to solve S

SNCOS60 = 60N
SN = 60/cos60
SN =120N

Try to solve R

RN + 120sin60 = 80N
RN = 80 - 120sin60
RN = -23.92

Can anybody spot a mistake? I'm sure you use SIN for forces up and down and COS for forces horizontally.
• December 29th 2008, 10:01 AM
djmccabie
looking at it again would i find RN using 120sin30 ?
• December 29th 2008, 06:52 PM
skeeter
what diagram?
• December 30th 2008, 02:35 AM
djmccabie
Its this 1

http://www.mathhelpforum.com/math-he...1&d=1230633116

seem this site doesnt host pictures for long lol
• December 30th 2008, 02:41 AM
mr fantastic
Quote:

Originally Posted by djmccabie
Its this 1

http://www.mathhelpforum.com/math-he...1&d=1230633116

seem this site doesnt host pictures for long lol

$S \cos 30^0 = 60$ .... (1)

$R + S \sin 30^0 = 80$ .... (2)

Solve equations (1) and (2).
• December 30th 2008, 07:15 AM
skeeter
Quote:

Originally Posted by djmccabie
Its this 1

http://www.mathhelpforum.com/math-he...1&d=1230633116

seem this site doesnt host pictures for long lol

http://www.mathhelpforum.com/math-he...-question.html
• December 31st 2008, 07:41 AM
RossBrons
As all forces are acting at one point, and that the point is being held in equilibrium by these forces it is said that the opposite forces are of equal magnitude. Mr Fantastic has detailed this with the two equations for vertical and horzontal force.

Since

S = (60/cos30)

Then

2R = 160-S

This can be simplified to be....

R = 80 - (30/cos30)

Kind Regards,

Ross