# forces problem

• Apr 13th 2007, 04:10 AM
Mr_Green
forces problem
• Apr 13th 2007, 06:27 AM
earboth
Quote:

Originally Posted by Mr_Green

Hi, Mr_Green,

in the attachment you'll find a your slightly modified diagram.

You are dealing with a parallelogram of forces. Take the left half of the parallelogram. Since AD || BC you only have to calculate the side AD = BC. Use Sine rule:

AD/70 N = sin(45°)/sin(30°)

AD = 140*1/2*sqrt(2) N = 70*sqrt(2) N ≈ 98.994949.. N ≈ 99 N
• Apr 14th 2007, 04:43 AM
topsquark
Quote:

Originally Posted by Mr_Green

This is the more "traditional" way to do this:

This is a Newton's 2nd Law problem.

Let the tension on the left be T1 = 70 N and the tension on the right be T2. Let the mass be m. I will use a coordinate system with +x to the right and +y upward.

Now take Newton's 2nd Law in each coordinate direction acting on the point where the 3 strings meet:
(Sum)F_x = -70*cos(45) + T2*cos(60) = 0 <-- Since this point has no mass (and is not accelerating either.)

(Sum)F_y = 70*sin(45) + T2*sin(60) - mg = 0

From the x equation we know that:
T2 = 70*[cos(45)/cos(60)] N = 98.9949 N

As a bonus:

Inserting this into the second equation we get:
70*sin(45) + (98.9949)*sin(60) - mg = 0

m = 1/g*[70*sin(45) + (98.9949)*sin(60)] kg =
13.7989 kg

-Dan
• Apr 14th 2007, 10:22 PM
earboth
Quote:

Originally Posted by topsquark
...
m = 1/g*[70*sin(45) + (98.9949)*sin(60)] kg =
13?.7989 kg

-Dan

Hi, topsquark,

I've found the digit you've lost: ? = 5