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.

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Asked on July 9, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-09T17:30:25.480Z

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: F_1 = 5N along AB (from A to B) F_2 = 10N along BC (from B to C) F_3 = 8N along CD (from C to D) F_4 = 12N along AD (from A to D) F_5 = 6N along AC (from A to C) The length of the diagonal AC can be found using the Pythagorean theorem: AC = 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 × d, where F is the force and d 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. a) Find the sum of moments of these forces about A Step 1: Identify forces passing through A. Forces F_1 (5N along AB), F_4 (12N along AD), and F_5 (6N along AC) all pass through point A. Therefore, their moments about A are zero. M_AB,A = 0 M_AD,A = 0 M_AC,A = 0 Step 2: Calculate the moment of F_2 (10N along BC) about A. The line of action of F_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. M_BC,A = -10N × 4m = -40 Nm Step 3: Calculate the moment of F_3 (8N along CD) about A. The line of action of F_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. M_CD,A = -8N × 3m = -24 Nm Step 4: Sum all moments about A. M_A = M_AB,A + M_BC,A + M_CD,A + M_AD,A + M_AC,A M_A = 0 + (-40 Nm) + (-24 Nm) + 0 + 0 = -64 Nm The sum of moments about A is 64 Nm clockwise. b) Find the sum of moments of these forces about B Step 1: Identify forces passing through B. Forces F_1 (5N along AB) and F_2 (10N along BC) both pass through point B. Therefore, their moments about B are zero. M_AB,B = 0 M_BC,B = 0 Step 2: Calculate the moment of F_3 (8N along CD) about B. The line of action of F_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. M_CD,B = 8N × 3m = 24 Nm Step 3: Calculate the moment of F_4 (12N along AD) about B. The line of action of F_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. M_AD,B = -12N × 4m = -48 Nm Step 4: Calculate the moment of F_5 (6N along AC) about B. The line of action of F_5 is the diagonal AC. We need the perpendicular distance from B to the line AC. In right triangle ABC, the area is (1)/(2) × AB × BC = (1)/(2) × 4m × 3m = 6 m^2. The area can also be expressed as (1)/(2) × AC × h, where h is the perpendicular distance from B to AC. 6 m^2 = (1)/(2) × 5m × h h = (12)/(5)m = 2.4m The force acts from A to C. This creates a clockwise rotation about B. M_AC,B = -6N × 2.4m = -14.4 Nm Step 5: Sum all moments about B. M_B = M_AB,B + M_BC,B + M_CD,B + M_AD,B + M_AC,B M_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. c) 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). Step 1: Calculate the moment of F_1 (5N along AB) about E. The line of action of F_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. M_AB,E = 5N × 1m = 5 Nm Step 2: Calculate the moment of F_2 (10N along BC) about E. The line of action of F_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. M_BC,E = -10N × 4m = -40 Nm Step 3: Calculate the moment of F_3 (8N along CD) about E. The line of action of F_3 is the line y=3. The perpendicular distance from E(0,-1) to y=3 is 3 - (-1) = 4m. The force acts from C to D (leftwards). This creates a clockwise rotation about E. M_CD,E = -8N × 4m = -32 Nm Step 4: Calculate the moment of F_4 (12N along AD) about E. The line of action of F_4 is the line x=0. This line passes through E(0,-1). Therefore, its moment about E is zero. M_AD,E = 0 Step 5: Calculate the moment of F_5 (6N along AC) about E. The line of action of F_5 is the line AC, which passes through A(0,0) and C(4,3). The equation of this line is y = (3)/(4)x, or 3x - 4y = 0. The perpendicular distance from E(0,-1) to 3x - 4y = 0 is: d = (|3(0) - 4(-1)|)/(sqrt(3^2 + (-4)^2)) = (|4|)/(sqrt(9+16)) = (4)/(sqrt(25)) = (4)/(5)m = 0.8m The force acts from A to C. E is below A. This creates a clockwise rotation about E. M_AC,E = -6N × 0.8m = -4.8 Nm Step 6: Sum all moments about E. M_E = M_AB,E + M_BC,E + M_CD,E + M_AD,E + M_AC,E M_E = 5 Nm + (-40 Nm) + (-32 Nm) + 0 + (-4.8 Nm) M_E = 5 - 40 - 32 - 4.8 = -71.8 Nm The sum of moments about E is 71.8 Nm clockwise. Drop the next question.

Accepted Answer · 2026-07-09T17:30:25.480Z

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: F_1 = 5N along AB (from A to B) F_2 = 10N along BC (from B to C) F_3 = 8N along CD (from C to D) F_4 = 12N along AD (from A to D) F_5 = 6N along AC (from A to C) The length of the diagonal AC can be found using the Pythagorean theorem: AC = 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 × d, where F is the force and d 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. a) Find the sum of moments of these forces about A Step 1: Identify forces passing through A. Forces F_1 (5N along AB), F_4 (12N along AD), and F_5 (6N along AC) all pass through point A. Therefore, their moments about A are zero. M_AB,A = 0 M_AD,A = 0 M_AC,A = 0 Step 2: Calculate the moment of F_2 (10N along BC) about A. The line of action of F_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. M_BC,A = -10N × 4m = -40 Nm Step 3: Calculate the moment of F_3 (8N along CD) about A. The line of action of F_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. M_CD,A = -8N × 3m = -24 Nm Step 4: Sum all moments about A. M_A = M_AB,A + M_BC,A + M_CD,A + M_AD,A + M_AC,A M_A = 0 + (-40 Nm) + (-24 Nm) + 0 + 0 = -64 Nm The sum of moments about A is 64 Nm clockwise. b) Find the sum of moments of these forces about B Step 1: Identify forces passing through B. Forces F_1 (5N along AB) and F_2 (10N along BC) both pass through point B. Therefore, their moments about B are zero. M_AB,B = 0 M_BC,B = 0 Step 2: Calculate the moment of F_3 (8N along CD) about B. The line of action of F_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. M_CD,B = 8N × 3m = 24 Nm Step 3: Calculate the moment of F_4 (12N along AD) about B. The line of action of F_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. M_AD,B = -12N × 4m = -48 Nm Step 4: Calculate the moment of F_5 (6N along AC) about B. The line of action of F_5 is the diagonal AC. We need the perpendicular distance from B to the line AC. In right triangle ABC, the area is (1)/(2) × AB × BC = (1)/(2) × 4m × 3m = 6 m^2. The area can also be expressed as (1)/(2) × AC × h, where h is the perpendicular distance from B to AC. 6 m^2 = (1)/(2) × 5m × h h = (12)/(5)m = 2.4m The force acts from A to C. This creates a clockwise rotation about B. M_AC,B = -6N × 2.4m = -14.4 Nm Step 5: Sum all moments about B. M_B = M_AB,B + M_BC,B + M_CD,B + M_AD,B + M_AC,B M_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. c) 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). Step 1: Calculate the moment of F_1 (5N along AB) about E. The line of action of F_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. M_AB,E = 5N × 1m = 5 Nm Step 2: Calculate the moment of F_2 (10N along BC) about E. The line of action of F_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. M_BC,E = -10N × 4m = -40 Nm Step 3: Calculate the moment of F_3 (8N along CD) about E. The line of action of F_3 is the line y=3. The perpendicular distance from E(0,-1) to y=3 is 3 - (-1) = 4m. The force acts from C to D (leftwards). This creates a clockwise rotation about E. M_CD,E = -8N × 4m = -32 Nm Step 4: Calculate the moment of F_4 (12N along AD) about E. The line of action of F_4 is the line x=0. This line passes through E(0,-1). Therefore, its moment about E is zero. M_AD,E = 0 Step 5: Calculate the moment of F_5 (6N along AC) about E. The line of action of F_5 is the line AC, which passes through A(0,0) and C(4,3). The equation of this line is y = (3)/(4)x, or 3x - 4y = 0. The perpendicular distance from E(0,-1) to 3x - 4y = 0 is: d = (|3(0) - 4(-1)|)/(sqrt(3^2 + (-4)^2)) = (|4|)/(sqrt(9+16)) = (4)/(sqrt(25)) = (4)/(5)m = 0.8m The force acts from A to C. E is below A. This creates a clockwise rotation about E. M_AC,E = -6N × 0.8m = -4.8 Nm Step 6: Sum all moments about E. M_E = M_AB,E + M_BC,E + M_CD,E + M_AD,E + M_AC,E M_E = 5 Nm + (-40 Nm) + (-32 Nm) + 0 + (-4.8 Nm) M_E = 5 - 40 - 32 - 4.8 = -71.8 Nm The sum of moments about E is 71.8 Nm clockwise. Drop the next question.

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.

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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.

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