- #1

IDK10

- 67

- 3

## Homework Statement

We are given a diagram to which we need to make measurements of. There is a line Y which is 9.5cm long. At one end, point X, is a force of 6,500N going perpendicular to the line; at the other end is point J, which the line continues for another 4.5cm to point P which is the pivot point.

At point J, there is a line going 60° towards point P, this is another force line, it is going on the same general direction as the other force line. All lines, except force lines are at a 1:10 scale. Using the measurements, we need to find the unknown force of the second force line.

## Homework Equations

I know the sum of counterclockwise moments equal the sum of clockwise moments. Also, if it is at an angle, you use sin(thetha)x the length of the line to the pivot which is perpendicular to the line of action of the force.

## The Attempt at a Solution

I tried find the moment of the kniwn force, 6,500 x (9.5x10)/100= 910Nm. Then I used 910 + w

_{2}x

_{2}= 0

w

_{2}0.21 [From 2.1x(100/10). Also, I got 2.1 from measuring the length of the line from P which meets up with the line perpendicular to that of the force]

_{2}= -910

-910/0.21=4,333.333... (I know its meant to be negative, but I didn't think a negative force would work)