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Math Help - Some Random Questions....

  1. #1
    Member classicstrings's Avatar
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    Some Random Questions....

    Hello! I'm having some difficulty with these questions. Thanks for the help!

    1.

    If a = 2j - 3k
    b = 3i - 3j + ck

    and

    abs(a+b) = root35

    Then c is equal?

    2.

    The angle between vectors a and b is 120 and abs of a = abs of b. The angle b/w a and a + b is?

    3.

    A woman of mass 50kg stands in a lift which is moving upwards with acceleration of am/s^2. THe floor of the lift exerts a force of magnitude 75g Newtons on the women. Then value of a is?

    4.

    A body of mass 15kg slides down a straight slide which is inclined at an angle of 30 degrees to horizontal. The normal force, in Newtons of the slide acting on the boy is?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by classicstrings View Post
    1.

    If a = 2j - 3k
    b = 3i - 3j + ck

    and

    abs(a+b) = root35

    Then c is equal?
    a+b=3i + (2-3)j + (-3+c)k = 3i -j+(c-3)k

    so

    |a+b|^2 = 3^2 + 1 + (c-3)^2 = c^2 - 6c + 19

    so if |a+b|=sqrt(35) then:

    c^2 - 6c + 19 = 35,

    or:

    c^2 - 6c -16 = 0

    which factorises by inspection to:

    (c-8)(c+2)=0,

    so c=8 or -2.

    RonL
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  3. #3
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    Hello, classicstrings!

    I drew a sketch for #2 . . . and saw an "eyeball" solution!


    \text{2. The angle between vectors }\vec{a}\text{ and }\vec{b}\text{ is }120^o,\;\text{ and }|a| = |b|.
    \text{Find the angle between }\vec{a}\text{ and }\overrightarrow{a + b}
    Code:
          *
           *  *
            *     *
           b *        * a+b
              *           *
               * 120*         *
                * * * * * * * * * *
                         a

    We are given: The angle between \vec{a} and \vec{b} is 120^o and |a| = |b|.

    We have an isosceles triangle with equal sides |a| and vertex angle 120^o.

    Therefore, the base angles are 30^o.

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    3. A woman of mass 50kg stands in a lift which is moving upwards with acceleration of am/s^2. THe floor of the lift exerts a force of magnitude 75g Newtons on the women. Then value of a is?
    When in doubt, use Newton's 2nd. I have a Free-Body Diagram for the woman. There is a normal force, N, acting upward, she has a weight, w = mg, acting downward. The lift the woman is standing in is accelerating upward at a m/s^2, so she is as well. I have a +y direction in the upward direction.

    The floor of the lift is exerting a force of 75g N on her. This is the normal force, so N = 75g N. Thus Newton's 2nd on the woman reads:
    \sum F_y = N - w = ma

    75g - mg = ma

    a = \frac{1}{m} (75g - mg) where m = 50 kg.

    Thus a = 4.9 m/s^2.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    4. A body of mass 15kg slides down a straight slide which is inclined at an angle of 30 degrees to horizontal. The normal force, in Newtons of the slide acting on the boy is?
    I have a Free-Body Diagram of the boy. There is a normal force, N, acting perpendicularlly out of the slide, there is a weight, w = mg, acting straight down. Friction is not mentioned in the problem so I will ignore it. I am going to set a +x direction down the slide and a +y direction in the direction of the normal force.

    Newton's 2nd in the +y direction says:
    \sum F_y = N - w \cdot cos(30) = ma_y <--Make sure you understand why this is cosine, not sine!

    There is no acceleration in the y direction so a_y = 0.

    N - mg \cdot cos(30) = 0

    N = mg \cdot cos(30) = 127.306 \, N

    or N = 130 N (keeping 2 significant figures.)

    -Dan
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