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Sliding ladder leaning against wall, and a triangle of maximum area

  1. Jul 5, 2005 #1
    Hello,
    Below are two questions i am unsure of. :uhh: Can someone please verify my answers and tell me if i am doing something wrong. :yuck: thank-you for your help. :smile:

    1. A rigid beam 30 m long is leaning against a vertical wall. If the bottom of the beam is pulled horizontally away from the wall at 3 m/s, how fast is the angle between the beam and the ground changing when the bottom of the beam is 18 m from the wall?

    sinθ = y/30

    cosθ = 18/30
    θ= 53.13

    A= bh/2
    A= [(x)(30sinθ)] / 2
    A'= [(30sinθ) (dx/dt) + (x)(30cosθ) (dθ/dt)] / 2
    0= [30sin(53.13) (3) + (18)(30cos53.13) (dθ/dt)] / 2
    dθ/dt= -6.67 m/s

    2.Triangle ABC is inscribed in a semicircle with diameter BC=12 cm. Find the value of angle B that produces the triangle of maximum area.
    Hint: An angle inscribed in a semi-circle is a right triangle.

    A=bh/2
    A=(12cosB)(12sinB)
    =72(2cosBsinB)
    =72(sin2B)
    A'=72(cos2B)
    0=72cos2B
    cos2B=0
    2B=90°
    B=45°

    i got sin2B using the trig double angle forumla sin2x=2sinxcosx
     
  2. jcsd
  3. Jul 5, 2005 #2

    dextercioby

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    There's a very elegant approach for the second problem. No calculus required (!) :surprised:

    [tex] A_{\mbox{triangle}} =\frac{AB\cdot AC}{2} [/tex] (1)

    Constraint, following Pythagora's theorem

    [tex] AB^{2}+AC^{2}=12^{2} [/tex] (2)

    (2) can be written

    [tex] \left(AB-AC\right)^{2} =12^{2}-2 AB\cdot AC [/tex] (2')

    U impose that the product (which by (1)is ~Area) be maximum, therefore the RHS would have to be minimum. Since the LHS can be no less than 0, one gets that it has to be 0.

    [tex] AB=AC \Rightarrow \hat{CBA} =45 \mbox{deg} [/tex]

    Daniel.
     
  4. Jul 5, 2005 #3
    thanks...but is the first question right?
     
  5. Jul 5, 2005 #4

    dextercioby

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    The reasoning is okay. The area of the triangle is indeed constant. I didn't check the numbers. I hope they're okay.

    Daniel.
     
  6. Jul 5, 2005 #5
    You forgot the chain rule here, but in the end it didn't affect your answer.

    [tex]A = 72\sin{(2B)}[/tex]
    [tex]A' = 72\cos{(2B)}*2[/tex]
     
  7. Jul 6, 2005 #6

    HallsofIvy

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    Are you referring to problem 1? Why would the area be constant? The area goes to 0 as the beam falls to the floor. What is definitely constant is the length of the beam- the length of the hypotenuse: Pythagorean theorem: x2+ y2= 900 where x is the distance of the bottom of the beam from the wall and y is the height of the other end up the wall. Differentiating with respect to t,
    2xx'+ 2yy'= 0. At the given instant, x= 18 so y= [tex]\sqrt{900-18^2}= \sqrt{576}= 24. 18(3)+ 24(y')= 0 so y'= -54/24= -2.25. y= 30 sin(θ) so
    y'= 30 cos(θ)θ'. When x= 18, &theta= 53.13 degrees and
    cos(53.13)= 0.6000. -2.25= (30)(0.6)(θ') so θ'= (-2.25)/(18)= -.125 or -1/8 degrees per second.
     
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