Lever Arm
In the lever arm below, calculate the moment force (F) causes on Point A.
![](/img/v3/74ef08b5-f578-4396-9b05-1b8efd0d1930.png)
Expand Hint
The diagonal force applied at the end of the lever arm can be broken down into a
$$F_x$$
component.
$$$F_x=F\cdot cos(\theta)$$$
Hint 2
The diagonal force applied at the end of the lever arm can be broken down into a
$$F_y$$
component.
$$$F_y=F\cdot sin(\theta)$$$
The diagonal force applied at the end of the lever arm can be broken down into
$$F_x$$
and
$$F_y$$
components.
$$$F_x=F\cdot cos(\theta)=700\cdot cos(30)=606.22\:N$$$
$$$F_y=F\cdot sin(\theta)=700\cdot sin(30)=350\:N$$$
![](/img/v3/cc246259-cb63-4964-9af9-e166c2dd7e3a.png)
A system of two forces that are equal in magnitude, opposite in direction, and parallel to each other is called a couple. A moment (M) is defined as the cross product of the radius vector (r) and the force (f) from a point to the force's line of action.
$$$M=r \times F$$$
$$$M_z=xF_y-yF_x$$$
Thus, the moment about Point A:
$$$M_A=(6m)(350N)-(1m)(606.22N)=2100-606.22=1494\:N\cdot m$$$
$$$1494\:N\cdot m$$$
Time Analysis
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