Asked by audrey anne
a seesaw consists of a plank 4.5m long which is supported by a pivot at its center and moves in a vertical plane above the pivot. if the height of the pivot pillar above the ground is 1m, through what maximum angle can the seesaw beam move?
Answers
Answered by
Reiny
Did you make a sketch ?
I see an isosceles triangle on its side with equal sides 4.5 and a vertical line of 1
label the top of the plank P, the fulcrum F
draw a horizontal to hit the height at Q
PFQ is a right-angled triangle with hypotenuse 4.5, and a vertical leg of .5
angle QFP is half of the angle we want.
tan QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.34°
So the beam moves through an angle of 12.68°
I see an isosceles triangle on its side with equal sides 4.5 and a vertical line of 1
label the top of the plank P, the fulcrum F
draw a horizontal to hit the height at Q
PFQ is a right-angled triangle with hypotenuse 4.5, and a vertical leg of .5
angle QFP is half of the angle we want.
tan QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.34°
So the beam moves through an angle of 12.68°
Answered by
Reiny
we could have just used the cosine law
1^2 = 4.5^2 + 4.5^2 - 2(4.5)(4.5)cosØ
1 = 20.25 + 20.25 - 40.5cosØ
40.5 cosØ = 39.5
cosØ = 39.5/40.5 = .9753
Ø = 12.759°
And I just realized that in my first solution I should have used the sine instead of tangent
<b>sin QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.3794°
So the beam moves through an angle of 12.759° </b>
just like above
1^2 = 4.5^2 + 4.5^2 - 2(4.5)(4.5)cosØ
1 = 20.25 + 20.25 - 40.5cosØ
40.5 cosØ = 39.5
cosØ = 39.5/40.5 = .9753
Ø = 12.759°
And I just realized that in my first solution I should have used the sine instead of tangent
<b>sin QFP = .5/4.5 = 5/45 = 1/9
angle QFP = appr 6.3794°
So the beam moves through an angle of 12.759° </b>
just like above
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