Lever arm rule referrers to the similarity of triangles (similar triangles are the ones which have corresponding equal angles). They have the property of having the same ratio between the corresponding edges. This is in fact the lever-arm rule: the ratio between the lengths of the corresponding edges of two similar triangles are the same. By "corresponding" edges, one understands the edges of the triangles which oppose congruent angles.
For exemplification, let's take a triangle OAB. Drawing a parallel to AB which intersects OA and OB in A', respectively B', another triangle OA'B' is formed. The two triangles OAB and OA'B' are similar. Then,
[OA']/[OA]=[A'B']/[AB]=[OB']/[OB]
Arm is the example of 2nd class lever.
Arm is the example of 2nd class lever.
no because to get a torque you must multiply lever arm by force. If lever is zero, then torque is zero
1.25
first class lever
Arm is the example of 2nd class lever.
Arm is the example of 2nd class lever.
no because to get a torque you must multiply lever arm by force. If lever is zero, then torque is zero
Yes!
The end of a lever that carries the load is the output arm instead of the input arm which is the end of a lever that force is applied to move the load.
1.25
first class lever
No it isn't. It is a third class lever.
Divide the length of the force arm by the length of the resistance arm.
Your arm
arm
a lever