A seesaw consisting of a uniform board of mass M and length L, supports at rest a father and daughter with masses "mf " and "md", respectively. The support ( called the fulcrum ) is under the centre of gravity of the board , the father is the distance "d" from the center and the daughter is a distance "L/2" from the center.

A) determine the magnitude of the upward force "n" exerted by the support on the board ?

B) determine where the father should sit to balance the system at rest ?

A) N = (M + mf + md) g

B) d = md L / (2 mf)

Explanation:

Draw a free body diagram.  There are four forces acting on the board.

Weight force M g pulling down at the center of the board.

Normal force N pushing up at the center of the board.

Weight force mf g pulling down a distance d from the center.

Weight force md g pulling down at the end of the board.

Sum of forces in the y direction:

∑F = ma

N − M g − mf g − md g = 0

N = (M + mf + md) g

Sum of moments about the center of the board:

∑τ = Iα

md (L/2) − mf d = 0

d = md L / (2 mf)

The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = 9.8M Newton

The father should sit at L/4 at the other end in order to balance the system at rest

Since the seesaw is of a uniform board, the center of gravity will act at the center. If it is of length L, the support called the fulcrum will be positioned at L/2.

If the daughter is a distance L/2 from the center, Then, she must be positioned at L/4 from the one edge of the board.

A. The magnitude of the upward force n exerted by the support on the board will be the same with the magnitude of the weight of the seesaw. Which is

W = mg

W = 9.8M Newton

The weight of both the father and the daughter will act downward. so, they will not be included.

The father should sit at L/4 at the other end in order to balance the system at rest

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