# Homework Help: Forces necessary to hold a board steady (double check my work?)

1. Nov 25, 2012

### Lotus93

This is the problem:
http://oi47.tinypic.com/6qij4o.jpg

I set fx, fy, and B equal to zero.
'x' forces: Fx - Bsin(30) = 0
'y' forces: Fy - mg + B(cos(30)) = 0
lower end: -0.5*mg(cos(30)) + 0.4*B = 0

Using the third equation I solved for B, and got -477.4.
I then plugged B into the first two equations and calculated Fy=854.4 and Fx=-238.7.

I don't feel confident in these numbers, so I wanted to double check with someone who has a better understanding. Did I go wrong somewhere, or does this look okay?

2. Nov 25, 2012

### SammyS

Staff Emeritus
Why set all those to zero?
B should be positive !

3. Nov 26, 2012

### Lotus93

I set them equal to zero because I thought the net force should be equal to 0 in order to hold the board still... is that wrong?

4. Nov 26, 2012

### Lotus93

This is what I have now...

(-0.5)(m)(g)(cos(30))+(B)(0.4)=0
B=-(-.5)(45)(9.8)(cos(30))/0.4
B= 477.4

Fx-B(sin(30))=0
Fx=(477.4)(sin(30))
Fx= 238.7

Fy-mg+B(Cos(30))=0
Fy=-477.4(Cos(30))+(45)(9.8)
Fy=27.6

Am I anywhere close to being right?

Last edited: Nov 26, 2012
5. Nov 26, 2012

### SammyS

Staff Emeritus
The net force is zero, but that doesn't mean that each individual component making up the net force is zero.

6. Nov 26, 2012

### SammyS

Staff Emeritus
The method is right.

The numbers look reasonable.

Of course, all of those answers are in units of Newtons.