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ABCD is a rectangle where AB = 4m and BC = 3m. Forces 5N, 10N, 8N, 12N and 6N acts along the lines AB, BC, CD, AD and AC respectively in the order of letters. Find the sum of moments of these forces about: a) A, b) B, c) E where C is on DA produced and AE = 1m.

Mathematics

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3 parts
a

64 Nm clockwise\text{64 Nm clockwise}

b

38.4 Nm clockwise\text{38.4 Nm clockwise}

c

71.8 Nm clockwise\text{71.8 Nm clockwise}

Step-by-step solution

15 steps

First, let's define the rectangle ABCD and the forces acting on it.
Given: ABCD is a rectangle with AB = 4m and BC = 3m.
Therefore, CD = 4m and AD = 3m.
Let's set up a coordinate system with A at the origin (0,0).
A = (0,0)
B = (4,0)
C = (4,3)
D = (0,3)

The forces and their lines of action are:

  • F1=5NF_1 = 5N along AB (from A to B)
  • F2=10NF_2 = 10N along BC (from B to C)
  • F3=8NF_3 = 8N along CD (from C to D)
  • F4=12NF_4 = 12N along AD (from A to D)
  • F5=6NF_5 = 6N along AC (from A to C)

The length of the diagonal AC can be found using the Pythagorean theorem:

AC=AB2+BC2=42+32=16+9=25=5mAC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5m

We will use the convention that counter-clockwise (CCW) moments are positive and clockwise (CW) moments are negative. The moment of a force about a point is given by M=F×dM = F \times d, where FF is the force and dd is the perpendicular distance from the point to the line of action of the force. If a force's line of action passes through the point, its moment about that point is zero.

Part a· 4 steps

Find the sum of moments of these forces about A

  1. Identify forces passing through A.

    Forces F1F_1 (5N along AB), F4F_4 (12N along AD), and F5F_5 (6N along AC) all pass through point A. Therefore, their moments about A are zero.

    MAB,A=0M_{AB,A} = 0 MAD,A=0M_{AD,A} = 0 MAC,A=0M_{AC,A} = 0

  2. Calculate the moment of F2F_2 (10N along BC) about A.

    The line of action of F2F_2 is the line x=4. The perpendicular distance from A(0,0) to this line is AB = 4m.
    The force acts from B to C (upwards). This creates a clockwise rotation about A.

    MBC,A=−10N×4m=−40NmM_{BC,A} = -10N \times 4m = -40 Nm

  3. Calculate the moment of F3F_3 (8N along CD) about A.

    The line of action of F3F_3 is the line y=3. The perpendicular distance from A(0,0) to this line is AD = 3m.
    The force acts from C to D (leftwards). This creates a clockwise rotation about A.

    MCD,A=−8N×3m=−24NmM_{CD,A} = -8N \times 3m = -24 Nm

  4. Sum all moments about A.

    MA=MAB,A+MBC,A+MCD,A+MAD,A+MAC,AM_A = M_{AB,A} + M_{BC,A} + M_{CD,A} + M_{AD,A} + M_{AC,A} MA=0+(−40Nm)+(−24Nm)+0+0=−64NmM_A = 0 + (-40 Nm) + (-24 Nm) + 0 + 0 = -64 Nm

    The sum of moments about A is 64 Nm clockwise\boxed{\text{64 Nm clockwise}}.

Part b

38.4 Nm clockwise\text{38.4 Nm clockwise}

Find the sum of moments of these forces about B

  1. Identify forces passing through B.

    Forces F1F_1 (5N along AB) and F2F_2 (10N along BC) both pass through point B. Therefore, their moments about B are zero.

    MAB,B=0M_{AB,B} = 0 MBC,B=0M_{BC,B} = 0

  2. Calculate the moment of F3F_3 (8N along CD) about B.

    The line of action of F3F_3 is the line y=3. The perpendicular distance from B(4,0) to this line is BC = 3m.
    The force acts from C to D (leftwards). This creates a counter-clockwise rotation about B.

    MCD,B=8N×3m=24NmM_{CD,B} = 8N \times 3m = 24 Nm

  3. Calculate the moment of F4F_4 (12N along AD) about B.

    The line of action of F4F_4 is the line x=0. The perpendicular distance from B(4,0) to this line is AB = 4m.
    The force acts from A to D (upwards). This creates a clockwise rotation about B.

    MAD,B=−12N×4m=−48NmM_{AD,B} = -12N \times 4m = -48 Nm

  4. Calculate the moment of F5F_5 (6N along AC) about B.

    The line of action of F5F_5 is the diagonal AC. We need the perpendicular distance from B to the line AC.
    In right triangle ABC, the area is 12×AB×BC=12×4m×3m=6m2\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 4m \times 3m = 6 m^2.
    The area can also be expressed as 12×AC×h\frac{1}{2} \times AC \times h, where hh is the perpendicular distance from B to AC.

    6m2=12×5m×h  ⟹  h=125m=2.4m6 m^2 = \frac{1}{2} \times 5m \times h \implies h = \frac{12}{5}m = 2.4m

    The force acts from A to C. This creates a clockwise rotation about B.

    MAC,B=−6N×2.4m=−14.4NmM_{AC,B} = -6N \times 2.4m = -14.4 Nm

  5. Sum all moments about B.

    MB=MAB,B+MBC,B+MCD,B+MAD,B+MAC,BM_B = M_{AB,B} + M_{BC,B} + M_{CD,B} + M_{AD,B} + M_{AC,B} MB=0+0+24Nm+(−48Nm)+(−14.4Nm)=24−48−14.4=−38.4NmM_B = 0 + 0 + 24 Nm + (-48 Nm) + (-14.4 Nm) = 24 - 48 - 14.4 = -38.4 Nm

    The sum of moments about B is 38.4 Nm clockwise\boxed{\text{38.4 Nm clockwise}}.

Part c

71.8 Nm clockwise\text{71.8 Nm clockwise}

Find the sum of moments of these forces about E where C is on DA produced and AE = 1m.

The phrase "C is on DA produced" is likely a typo. We will interpret "E is on DA produced and AE = 1m" to mean that E is on the line containing AD, extended past A, such that the distance AE is 1m.
Since A is (0,0) and D is (0,3), the line AD is the y-axis (x=0). Extending past A means E is at (0,-1).
E = (0,-1).

  1. Calculate the moment of F1F_1 (5N along AB) about E.

    The line of action of F1F_1 is the line y=0. The perpendicular distance from E(0,-1) to y=0 is 1m.
    The force acts from A to B (rightwards). This creates a counter-clockwise rotation about E.

    MAB,E=5N×1m=5NmM_{AB,E} = 5N \times 1m = 5 Nm

  2. Calculate the moment of F2F_2 (10N along BC) about E.

    The line of action of F2F_2 is the line x=4. The perpendicular distance from E(0,-1) to x=4 is 4m.
    The force acts from B to C (upwards). This creates a clockwise rotation about E.

    MBC,E=−10N×4m=−40NmM_{BC,E} = -10N \times 4m = -40 Nm

  3. Calculate the moment of F3F_3 (8N along CD) about E.

    The line of action of F3F_3 is the line y=3. The perpendicular distance from E(0,-1) to y=3 is 3−(−1)=4m3 - (-1) = 4m.
    The force acts from C to D (leftwards). This creates a clockwise rotation about E.

    MCD,E=−8N×4m=−32NmM_{CD,E} = -8N \times 4m = -32 Nm

  4. Calculate the moment of F4F_4 (12N along AD) about E.

    The line of action of F4F_4 is the line x=0. This line passes through E(0,-1). Therefore, its moment about E is zero.

    MAD,E=0M_{AD,E} = 0

  5. Calculate the moment of F5F_5 (6N along AC) about E.

    The line of action of F5F_5 is the line AC, which passes through A(0,0) and C(4,3). The equation of this line is y=34xy = \frac{3}{4}x, or 3x−4y=03x - 4y = 0.
    The perpendicular distance from E(0,-1) to 3x−4y=03x - 4y = 0 is:

    d=∣3(0)−4(−1)∣32+(−4)2=∣4∣9+16=425=45m=0.8md = \frac{|3(0) - 4(-1)|}{\sqrt{3^2 + (-4)^2}} = \frac{|4|}{\sqrt{9+16}} = \frac{4}{\sqrt{25}} = \frac{4}{5}m = 0.8m

    The force acts from A to C. E is below A. This creates a clockwise rotation about E.

    MAC,E=−6N×0.8m=−4.8NmM_{AC,E} = -6N \times 0.8m = -4.8 Nm

  6. Sum all moments about E.

    ME=MAB,E+MBC,E+MCD,E+MAD,E+MAC,EM_E = M_{AB,E} + M_{BC,E} + M_{CD,E} + M_{AD,E} + M_{AC,E} ME=5Nm+(−40Nm)+(−32Nm)+0+(−4.8Nm)M_E = 5 Nm + (-40 Nm) + (-32 Nm) + 0 + (-4.8 Nm) ME=5−40−32−4.8=−71.8NmM_E = 5 - 40 - 32 - 4.8 = -71.8 Nm

    The sum of moments about E is 71.8 Nm clockwise\boxed{\text{71.8 Nm clockwise}}.

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