- #1

Sameh soliman

- 27

- 2

## Homework Statement

A load was put on the right side of a seesaw 0.6 meter from its center as shown, on the other side a force of 49N is applied perpendicular to the seesaw surface 1 meter from its center, the setup is tuned to an angle of 30 from the horizontal axis as shown.

On the right side there is a wall perpendicular to the ground surface and overlap the seesaw but having a gap in the middle to allow the seesaw to path throw it easily while the load is wider than the seesaw so that it can interact with the wall.

Neglecting all frictions state the following.

a) State if or if not the seesaw will be able to move and why.

b) State the total torque of the system.at its initial start

## Homework Equations

t = r x F[/B]

## The Attempt at a Solution

Below is my solution but i am not sure if it was right !

-First i assume the wall on the right side is not there

torque produced by the left side = 1m X 49N =49Nm

**ANTICLOCKWISE .**

torque produced on the right side is due to the vector force of gravity but only the sub-vector perpendicular to the seesaw surface which i figured that it will be =

cos30 X 5kg X 9.8m2 = 42.44N

so the torque on the right side = 0.6m x 42.44N = 25.46Nm CLOCKWISE

So the total torque of the system if the wall was not there will be =

49Nm - 25.46Nm = 23.54Nm ANTICLOCKWISE

-Now i study the interaction between the system and the wall

As the seesaw trying to rotate anticlockwise the object on the right will exert a force on the wall and since the total torque of the system is 23.54Nm so the total force the object can produce is =

F = t/r = 23.54Nm / 0.6 m = 39.23 Nm in a direction that is perpendicular to the seesaw surface .

but since there is no friction so the only force that will affect the wall is the sub-vector perpendicular to the wall surface which i figured out will be equal to =

cos60 x 39.23N = 19.62N

And since the wall is attached to the ground and will not move so the wall will exert a normal force on the system in the negative direction but only the sub vector that is perpendicular to the seesaw surface will affect the torque and i figured out that that sub vector =

cos60 x 19.62N = 9.81N in the negative direction

so the torque that this normal force will produce on the system =

0.6m x 9.81N = 5.89Nm CLOCKWISE

SO THE TOTAL TORQUE OF THE SYSTEM AT ITS INITIAL STAR WILL BE =

23.54Nm - 5.89Nm = 17.65Nm

AND SINCE THERE IS NO FRICTION SO THE SYSTEM WILL MOVE ANTICLOCKWISE WHILE THE OBJECT ON THE LEFT WILL SLIDE TOWARD THE CENTRE OF THE SEESAW

[/B]

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