Question

Q. A uniform seesaw , 5 m long is supported at its centre. A boy weighing 40 kgf sits at a distance of 1m from the centre of the seesaw.

(i). To which class of lever dose it belong ?

(ii). Find where a girl of weight 20 kgf must sits on the seesaw so as to balance the weight of the boy.

Q. Derive Relationship between mechanical advantage , Velocity ratio and efficiency of machine .​

Answer 1

Explanation:

i ). The seesaw has fulcrum F at its centre means between load and the effort. So it it Class I lever.

ii). To balance the weight , let the girl be sit at distance X meter from the centre on opposite Side :

So

Effort arm = X meter

Load Arm = 1 meter

Load given = 40 kgf

Effort = 20 kgf.

So :

Taking moment about F ,

40 ×1 = 20 × X

X = 40 /20 = 2 meter.

So girl must sit at dist 2 meter from center on opposite size of boy

Relationship between M.A , V.R and Efficiency is :

M.A = V.R × n.

M.A. is Mechanical Advantage

V.R. is Velocity Ratio

n is the efficiency of machine.

Other information refer in Attachment

Thank you

Answer 2

Explanation:

i it belong to class I lever.

ii. The girl sit should be X = 2 meter.

Relation is :

M.A = V.R × efficiency.