Questions

Here's a step-by-step solution to the problem, calculating the resultant force, its angle, and the location of the equivalent system.

Given Data:

• Forces: $F_1 = 283 \text{ N}$, $F_2 = 63 \text{ N}$
• Couple Moment: $M_C = 438 \text{ N} \cdot \text{m}$ (counter-clockwise)
• Angles: $\theta = 30^\circ$
• Dimensions: $a = 2 \text{ m}$, $b = 4 \text{ m}$, $c = 5 \text{ m}$

Points and Coordinates (assuming A is origin (0,0)):

• A = $(0, 0)$
• D = $(0, a) = (0, 2)$
• B = $(c, b) = (5, 4)$
• C = $(c, 0) = (5, 0)$ (This seems to be the corner where $F_2$ and $M_C$ are applied, based on the diagram)

1. Resolve Forces into Components:

• Force $F_1$: Acts at point D. The diagram indicates $F_1$ has components along the 3-4-5 triangle.

  • Horizontal component of $F_1$ ($F_{1x}$): $F_1 \times \frac{4}{5} = 283 \text{ N} \times \frac{4}{5} = 226.4 \text{ N}$ (acting to the left, so negative)
  • Vertical component of $F_1$ ($F_{1y}$): $F_1 \times \frac{3}{5} = 283 \text{ N} \times \frac{3}{5} = 169.8 \text{ N}$ (acting downwards, so negative)

• Force $F_2$: Acts at point C. It's at an angle $\theta = 30^\circ$ below the horizontal.

  • Horizontal component of $F_2$ ($F_{2x}$): $F_2 \cos \theta = 63 \text{ N} \times \cos(30^\circ) = 63 \times 0.8660 = 54.558 \text{ N}$ (acting to the left, so negative)
  • Vertical component of $F_2$ ($F_{2y}$): $F_2 \sin \theta = 63 \text{ N} \times \sin(30^\circ) = 63 \times 0.5 = 31.5 \text{ N}$ (acting upwards, so positive)

a) The magnitude of the resultant force FR: N

Sum of horizontal forces ($\sum F_x$): $\sum F_x = F_{1x} + F_{2x} = -226.4 \text{ N} - 54.558 \text{ N} = -280.958 \text{ N}$

Sum of vertical forces ($\sum F_y$): $\sum F_y = F_{1y} + F_{2y} = -169.8 \text{ N} + 31.5 \text{ N} = -138.3 \text{ N}$

Magnitude of the resultant force ($F_R$): $F_R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ $F_R = \sqrt{(-280.958)^2 + (-138.3)^2}$ $F_R = \sqrt{78937.1 + 19126.89}$ $F_R = \sqrt{98063.99}$ $F_R \approx 313.15 \text{ N}$

The magnitude of the resultant force FR is 313.15 N.

b) The angle of the resultant force with respect to horizontal, assume positive to be counter-clockwise. $\theta_R$:

The angle $\alpha$ with the negative x-axis (since both $\sum F_x$ and $\sum F_y$ are negative, the resultant is in the third quadrant): $\tan \alpha = \frac{|\sum F_y|}{|\sum F_x|} = \frac{138.3}{280.958} \approx 0.4922$ $\alpha = \arctan(0.4922) \approx 26.20^\circ$

The angle $\theta_R$ measured counter-clockwise from the positive x-axis: $\theta_R = 180^\circ + \alpha = 180^\circ + 26.20^\circ = 206.20^\circ$

The angle of the resultant force with respect to horizontal is 206.20 degrees.

c) The perpendicular distance of the vertical resultant force to the middle line of the vertical bar.

This part asks to replace the force system with an equivalent resultant force where the line of action intersects member BC (the horizontal bar) at a certain point. It also asks for the perpendicular distance of the vertical resultant force to the middle line of the vertical bar. This wording is a bit ambiguous. It seems to refer to locating the resultant with respect to the vertical member AB.

Let's find the moment of all forces and the couple moment about a convenient point, say point A (0,0), to find the location of the resultant.

Moment about A ($\sum M_A$):

• Moment due to $F_1$ components:

  • $F_{1x}$ passes through A, so no moment about A.
  • $F_{1y}$ causes a clockwise moment about A. Arm is the x-coordinate of D, which is 0. Wait, $F_{1y}$ acts at D(0,a). So $F_{1x}$ has a moment arm 'a' and $F_{1y}$ has a moment arm '0'. This means $F_{1y}$ has no moment about A, and $F_{1x}$ has a moment about A. Let's re-evaluate moments from point D.
  • The force $F_1$ acts at point D (0, a). $F_1$ has components $F_{1x} = -226.4 \text{ N}$ (left) and $F_{1y} = -169.8 \text{ N}$ (down).
  • Moment due to $F_{1x}$ about A: $F_{1x} \times a = (-226.4 \text{ N}) \times (2 \text{ m})$ (Counter-clockwise if $F_{1x}$ was to the right, but it's to the left and acts at y=a, so it's clockwise). $M_{F_{1x},A} = -226.4 \text{ N} \times 2 \text{ m} = -452.8 \text{ N} \cdot \text{m}$ (clockwise)
  • Moment due to $F_{1y}$ about A: $F_{1y}$ passes through the y-axis, so its moment arm is 0 with respect to A. $M_{F_{1y},A} = 0$.

• Moment due to $F_2$ components (acting at C(5,0)):

  • $F_{2x} = -54.558 \text{ N}$ (left). Arm is y-coordinate of C, which is 0. So $M_{F_{2x},A} = 0$.
  • $F_{2y} = 31.5 \text{ N}$ (upwards). Arm is x-coordinate of C, which is $c = 5 \text{ m}$. This causes a clockwise moment. $M_{F_{2y},A} = -(31.5 \text{ N}) \times (5 \text{ m}) = -157.5 \text{ N} \cdot \text{m}$ (clockwise)

• Couple Moment $M_C = 438 \text{ N} \cdot \text{m}$ (counter-clockwise) $M_C = +438 \text{ N} \cdot \text{m}$

Total Moment about A ($\sum M_A$): $\sum M_A = M_{F_{1x},A} + M_{F_{1y},A} + M_{F_{2x},A} + M_{F_{2y},A} + M_C$ $\sum M_A = -452.8 + 0 + 0 - 157.5 + 438$ $\sum M_A = -610.3 + 438 = -172.3 \text{ N} \cdot \text{m}$ (clockwise)

Now, let the line of action of the resultant force $F_R$ intersect the x-axis at $x_R$ and the y-axis at $y_R$. The moment of the resultant force about A is $M_A = \sum F_x \cdot y_R - \sum F_y \cdot x_R$. (Using the convention that forces cause moments based on their components and distances, or more simply, $M_A = R_y x_R - R_x y_R$).

Let's assume the question is asking for the x-intercept of the resultant force's line of action relative to the vertical member AB. If the resultant force were purely vertical, its line of action would be $x = x_R$. If it were purely horizontal, its line of action would be $y = y_R$. However, the resultant has both components.

The moment due to the resultant force about A is $\sum M_A = \sum F_y \cdot x_R - \sum F_x \cdot y_R$. But usually, we find the perpendicular distance from the moment center to the line of action of the resultant. Let $d$ be the perpendicular distance from A to the line of action of $F_R$. $|M_A| = F_R \times d$ $172.3 \text{ N} \cdot \text{m} = 313.15 \text{ N} \times d$ $d = \frac{172.3}{313.15} \approx 0.550 \text{ m}$

The question asks for "the perpendicular distance of the vertical resultant force to the middle line of the vertical bar". This phrasing is still confusing.

• "vertical resultant force" - this implies considering only the $\sum F_y$ component.
• "middle line of the vertical bar" - the vertical bar is from A to B. Its middle line is at $x=0$.
• This suggests we need the x-intercept of the resultant force.

Let the line of action of the resultant force be $y - y_0 = \tan \theta_R (x - x_0)$. The moment due to the resultant force about A (0,0) is: $\sum M_A = \sum F_x (y_0) - \sum F_y (x_0)$, where $(x_0, y_0)$ is any point on the line of action of $F_R$. Or, more commonly, $\sum M_A = -F_{Ry} \cdot x_{intercept} + F_{Rx} \cdot y_{intercept}$. We want to find $x_{intercept}$ (the point where the resultant crosses the x-axis). At this point, $y_{intercept} = 0$. So, $\sum M_A = -F_{Ry} \cdot x_{intercept}$ $-172.3 = -(-138.3) \cdot x_{intercept}$ $-172.3 = 138.3 \cdot x_{intercept}$ $x_{intercept} = \frac{-172.3}{138.3} \approx -1.246 \text{ m}$

This means the line of action of the resultant force crosses the x-axis (the line of the horizontal bar BC) at $x = -1.246$ m. This is to the left of point A.

The question asks for the perpendicular distance of the vertical resultant force to the middle line of the vertical bar. If the "vertical resultant force" means the entire resultant force $F_R$, and "middle line of the vertical bar" refers to the y-axis (the line $x=0$), then we are looking for the x-coordinate of the point where the line of action of $F_R$ crosses the x-axis, which is $x_{intercept}$. The distance from the "middle line of the vertical bar" ($x=0$) to the line of action of the vertical component of the resultant (if it were isolated) would be $x_{intercept}$.

Let's confirm the interpretation of "intersects member BC, measured from B". The line of action of the equivalent resultant force $F_R$ should have its moment about B (or any other point) equal to the total moment about B. Let's find $\sum M_B$: Point B is at $(c, b) = (5, 4)$. $\sum M_B = \sum M_A + \sum F_x \cdot y_B - \sum F_y \cdot x_B$ (This formula is for moving the moment center from A to B) $\sum M_B = -172.3 + (-280.958)(4) - (-138.3)(5)$ $\sum M_B = -172.3 - 1123.832 + 691.5$ $\sum M_B = -604.632 \text{ N} \cdot \text{m}$ (clockwise)

Now, the resultant force $F_R$ acting at $(x_R, y_R)$ must produce this moment about B. $F_R \cdot d_B = |M_B|$ where $d_B$ is the perpendicular distance from B. $d_B = \frac{604.632}{313.15} \approx 1.930 \text{ m}$.

If the line of action intersects member BC (the horizontal bar), it means it intersects the line $y=b=4$. Let the line of action be $y - y_0 = m(x - x_0)$, where $m = \tan \theta_R = \tan(206.20^\circ) = 0.4922$. We know $\sum M_A = \sum F_x (y_P) - \sum F_y (x_P)$ for any point $(x_P, y_P)$ on the line of action. $-172.3 = (-280.958)y_P - (-138.3)x_P$ $-172.3 = -280.958 y_P + 138.3 x_P$

We want to find the x-coordinate when $y_P=4$ (since it intersects member BC, which is at y=4). $-172.3 = -280.958(4) + 138.3 x_P$ $-172.3 = -1123.832 + 138.3 x_P$ $138.3 x_P = 1123.832 - 172.3$ $138.3 x_P = 951.532$ $x_P = \frac{951.532}{138.3} \approx 6.880 \text{ m}$

This is the x-coordinate where the line of action intersects the line $y=4$. The horizontal bar BC extends from $x=0$ (at B, if C is at (c,0) and B is at (c,b), but the diagram shows B at (c,b) which is (5,4) and C at (c,0) which is (5,0)). The diagram label BC is confusing. Let's assume the question meant member CB (the horizontal part of the L-frame). Point C is at (5,0) and B is at (5,4). Member CB is the vertical section $x=5$ from $y=0$ to $y=4$. The wording "intersects member BC" usually means the segment between points B and C.

Let's clarify points: A(0,0), D(0,a), C(c,0), B(c,b). So the vertical bar is from (0,0) to (0,b) (implied, by A and D and the structure). The horizontal bar is from (0,0) to (c,0) (implied) or from (c,0) to (c,b) (vertical). The diagram shows the vertical part AB and horizontal part BC. This implies A is (0,0), B is (c,0) = (5,0), and C is (c,b) = (5,4). This is consistent with the labels A, B, C on the diagram as points in space (A top left, B corner, C top right). Let's redefine coordinates based on this common interpretation for L-frames: A = (0, b) = (0,4) (top left corner) D = (0, b-a) = (0, 4-2) = (0,2) B = (c, b) = (5,4) (top right corner) C = (c, 0) = (5,0) (bottom right corner)

Let's re-calculate using these coordinates, as the initial assumption of A as origin is arbitrary and might not match the problem's intent for "vertical bar" and "member BC". It is more common for A to be a corner or a significant point. Given the diagram, A is the top-left corner, and B is the bottom-right corner of the L-frame. C is somewhere along the horizontal segment from B. D is on the vertical segment from A.

Let's assume the standard way to set up coordinates for this L-frame: origin at the corner where B is located, or A is top left. Let's try setting the origin at the point B (the corner where the two members meet). B = (0,0) A = $(-c, b) = (-5, 4)$ (This does not match given a,b,c. The diagram's a and b are vertical distances, and c is horizontal distance.)

Let's assume A is the origin (0,0) as usually done if not specified. A = (0,0) D = (0, a) = (0, 2) Let's assume the horizontal member extends from (0,0) to (c,0) and the vertical member extends from (0,0) to (0,b). This conflicts with the labels B and C on the diagram.

Let's use the explicit coordinate system implied by the problem statement and dimensions:

• Vertical bar: from A to some point below it. The label 'a' is a vertical distance from A to D. The label 'b' is a vertical distance from D to B. Total vertical length is $a+b = 2+4=6$ m. So A is at (0,6), D is at (0,4), and the vertical end is at (0,0).
• Horizontal bar: from the bottom of the vertical bar to C. The label 'c' is a horizontal distance from the vertical axis to C. So the horizontal end is at (c,0) = (5,0).

Re-evaluating points based on this common convention for L-frames (A at the top left):

• A = (0, $a+b$) = (0, 6)
• D = (0, b) = (0, 4) (This interpretation means 'a' is from A to D, 'b' is from D to B. The image implies 'a' is from A to D, and 'b' is from D to some point near B, then B is at the corner (c,0)?)

The labels 'a', 'b', 'c' are typical for dimensions, not coordinates from an implied origin. Let's stick to the most direct interpretation of the diagram's dimensions relative to the points:

• A is the top-left point.
• D is 'a' below A.
• The corner is labeled B. The vertical distance from D to B is 'b'. The horizontal distance from the vertical line (passing through A and D) to B is 'c'.
• C is somewhere on the horizontal line, possibly at the end of the beam or part of it. The diagram shows $F_2$ at point C.

Let's assume A is at (0, $y_A$). For convenience, let the origin be at the corner B. Let B = (0,0). Then:

• A = (0, $b+a$) = (0, $4+2$) = (0, 6)
• D = (0, $b$) = (0, 4)
• C = ($c$, 0) = (5, 0)

Now, recalculate forces and moments with B as origin:

Forces and Components (re-used):

• $F_1$: Applied at D (0, 4)

  • $F_{1x} = -226.4 \text{ N}$ (left)
  • $F_{1y} = -169.8 \text{ N}$ (down)

• $F_2$: Applied at C (5, 0)

  • $F_{2x} = -54.558 \text{ N}$ (left)
  • $F_{2y} = 31.5 \text{ N}$ (up)

Resultant Force ($F_R$): (This remains the same as it's independent of the origin choice) $F_R = 313.15 \text{ N}$ $\theta_R = 206.20^\circ$ (from positive x-axis counter-clockwise)

Moment about B ($\sum M_B$): (B is at origin (0,0))

• Couple Moment $M_C = 438 \text{ N} \cdot \text{m}$ (counter-clockwise)

• Moment due to $F_1$ components at D (0, 4):

  • $F_{1x} = -226.4 \text{ N}$ (left). Moment arm is y-coordinate of D (4m). Clockwise. $M_{F_{1x},B} = -(-226.4 \text{ N}) \times (4 \text{ m}) = 905.6 \text{ N} \cdot \text{m}$ (counter-clockwise, if force is left and point is above origin, it's positive $F_x \cdot y$) $M_{F_{1x},B} = -226.4 \times 4 = -905.6 \text{ N} \cdot \text{m}$ (clockwise)
  • $F_{1y} = -169.8 \text{ N}$ (down). Moment arm is x-coordinate of D (0m). $M_{F_{1y},B} = 0$.

• Moment due to $F_2$ components at C (5, 0):

  • $F_{2x} = -54.558 \text{ N}$ (left). Moment arm is y-coordinate of C (0m). $M_{F_{2x},B} = 0$.
  • $F_{2y} = 31.5 \text{ N}$ (up). Moment arm is x-coordinate of C (5m). Clockwise. $M_{F_{2y},B} = -(31.5 \text{ N}) \times (5 \text{ m}) = -157.5 \text{ N} \cdot \text{m}$ (clockwise)

Total Moment about B ($\sum M_B$): $\sum M_B = M_C + M_{F_{1x},B} + M_{F_{1y},B} + M_{F_{2x},B} + M_{F_{2y},B}$ $\sum M_B = 438 - 905.6 + 0 + 0 - 157.5$ $\sum M_B = 438 - 1063.1 = -625.1 \text{ N} \cdot \text{m}$ (clockwise)

c) The perpendicular distance of the vertical resultant force to the middle line of the vertical bar

Let the line of action of the resultant force be represented by $(x_R, y_R)$ relative to B (0,0). We know $\sum M_B = \sum F_y \cdot x_R - \sum F_x \cdot y_R$. $-625.1 = (-138.3) x_R - (-280.958) y_R$ $-625.1 = -138.3 x_R + 280.958 y_R$

The question asks for the perpendicular distance of the vertical resultant force to the middle line of the vertical bar. The "vertical bar" runs from A to B (along the y-axis in our current coordinate system). Its "middle line" is the y-axis itself, where $x=0$. A "vertical resultant force" means we are considering the resultant's effect only in terms of its vertical component's location. This implies finding the x-intercept of the line of action.

If we assume the line of action of the resultant is positioned such that its moment about B is $\sum M_B$, and we want to find its x-intercept (where it crosses the horizontal axis $y=0$). Let $y_R = 0$. $-625.1 = -138.3 x_R + 280.958(0)$ $-625.1 = -138.3 x_R$ $x_R = \frac{-625.1}{-138.3} \approx 4.520 \text{ m}$

This means the resultant force's line of action crosses the horizontal axis at $x=4.520$ m from B. The "middle line of the vertical bar" is the y-axis, $x=0$. The perpendicular distance from the y-axis to this point is $4.520 \text{ m}$.

It also states "where the line of action intersects member BC, measured from B". Member BC is the horizontal beam from B (0,0) to C (5,0). So, if the resultant intersects member BC, it means it intersects the x-axis ($y=0$). So the distance from B is $x_R = 4.520$ m. This is within the segment BC (0 to 5m).

Final Answers: a) The magnitude of the resultant force FR: 313.15 N b) The angle of the resultant force with respect to horizontal, assume positive to be counter-clockwise: 206.20 degrees c) The perpendicular distance of the vertical resultant force to the middle line of the vertical bar. This is interpreted as the x-intercept of the resultant's line of action, measured from the y-axis (the line of the vertical bar). Also, this point intersects member BC, measured from B. 4.520 m (to the right of B).