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Mathematics · May / June 2023 · Paper 51 · Question 4

YMA01

As printed in the paper

The question as text

4. A C D B 0.6 m 0.2 m 2 m Figure 1 Figure 1 shows a beam AB, of mass m kg and length 2 m, suspended by two light vertical ropes. The ropes are attached to the points C and D on the beam, where AC = 0.6 m and DB = 0.2 m The beam is in equilibrium in a horizontal position. A particle of mass pm kg is attached to the beam at A and the beam remains in equilibrium in a horizontal position. The beam is modelled as a uniform rod.

Read automatically from the paper; the printed page above is exact.

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  • 4(a)[7]

    Vertical resolution

    T

    4

    T

    pmg

    mg

    +

    =

    +

    OR a moments equation, see below.

    ( ) (

    ) (

    )

    (

    )

    M

    A

    : 4

    T

    ×

    0.6

    +

    T

    ×

    1.8

    =

    mg

    ×

    1

    Other moments equations:

    ( ) (

    ) (

    )

    (

    )

    M

    C

    :

    pmg

    ×

    0.6

    +

    T

    ×

    1.2

    =

    mg

    ×

    0.4

    ( ) (

    ) (

    )

    (

    )

    M

    G

    :

    pmg

    ×

    1

    +

    T

    ×

    0.8

    =

    4

    T

    ×

    0.4

    ( ) (

    ) (

    )

    (

    )

    M

    D

    :

    pmg

    ×

    1.8

    +

    mg

    ×

    0.8

    =

    4

    T

    ×

    1.2

    ( ) (

    ) (

    )

    (

    ) (

    )

    M

    B

    : 4

    T

    ×

    1.4

    +

    T

    ×

    0.2

    =

    pmg

    ×

    2

    +

    mg

    ×

    1

    Eliminate T

    5

    mg

    =

    5

    pmg

    +

    mg

    21

    4

    p =

    (exact ratio of 2 positive integers)

    21

    Tension at D is zero, seen or implied.

  • 4(b)[4]

    ( ) (

    )

    (

    )

    M

    C

    :

    qmg

    ×

    0.6

    =

    mg

    ×

    0.4

    .

    2

    q =

    (exact ratio of 2 positive integers), accept 0.666….. or 0.

    6

    3

    4(c) The centre of mass (or gravity) of the beam is in the middle; the mass

    (weight) of the beam acts at the middle, mass at centre, centre of mass at

    the centre. Penalise incorrect extras.

    Notes for Question 4

    (a)

    N.B. Full marks can be scored if consistent omission of g’s in a complete solution,

    but otherwise penalise omission of g’s

    B1

    Correct relationship between the tensions and placed correctly, seen or implied.

    M1

    Vertical resolution. Condone forces at C and D the wrong way round or written as

    T and

    T .

    C

    D

    This equation may be replaced with a moments equation.

    A1

    Correct unsimplified equation (even if T and 4T are the wrong way round on their

    diagram)

    M1

    Moments equation. Correct forces multiplied by a length. Condone consistent forces

    at C and D the wrong way round or written as

    T and

    T

    C

    D

    A1

    Correct unsimplified equation, in a variable consistent with their first equation.

    M1

    Eliminate T to give an equation in p only allow extra m’s or g’s or both

    A1

    Cao. Must be exact.

    N.B. If they write down more than two equations, award the marks for those

    equations which they use to solve the problem.

    (b)

    B1

    Recognise tension at D is 0, seen or implied

    M1

    Complete method to obtain an equation q only.

    e.g. Moments about C equation.

    A1

    Correct unsimplified equation in q only.

    A1

    Cao. Must be exact.

    ALT (b)

    M1

    Two other equations could be used and solved to find q .

    M0 if tension at D is never zero.

    A1

    Correct unsimplified equation in q only.

    A1

    Cao. Must be exact.

    Alternative equations:

    vert : '

    T

    =

    qmg

    +

    mg

    ( ) (

    )

    (

    )

    M

    A

    :

    T

    ' 0.6

    ×

    =

    mg

    ×

    1

    ( ) (

    )

    (

    )

    M

    G

    :

    qmg

    ×

    1

    =

    T

    ' 0.4

    ×

    ( ) (

    ) (

    )

    (

    )

    M

    D

    :

    qmg

    ×

    1.8

    +

    mg

    ×

    0.8

    =

    T

    ' 1.2

    ×

    ( ) (

    ) (

    )

    (

    )

    M

    B

    :

    qmg

    ×

    2

    +

    mg

    ×

    1

    =

    T

    ' 1.4

    ×

  • 4(c)[1]

    B1

    Any appropriate comment

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