Forces equal to P, 2P, 3P and 4P act along the sides of a square ABCD taken in order, find the magnitude, direction and the line of action of the resultant. [6]

Solution

Step 1: Define the coordinate system and assign forces

Let the square $ABCD$ have side length $a$. Place the square in a coordinate system such that:

• Point $A$ is at the origin $(0,0)$.
• Point $B$ is at $(a,0)$.
• Point $C$ is at $(a,a)$.
• Point $D$ is at $(0,a)$.

The forces act along the sides of the square in the following order:

• $P$ acts along $DA$ (upward, along the positive $y$-axis).
• $2P$ acts along $AB$ (rightward, along the positive $x$-axis).
• $3P$ acts along $BC$ (downward, along the negative $y$-axis).
• $4P$ acts along $CD$ (leftward, along the negative $x$-axis).

Step 2: Resolve forces into components

We resolve each force into its $x$- and $y$-components:

1. Force $P$ along $DA$ (upward):

   • $x$-component: $0$
   • $y$-component: $+P$

2. Force $2P$ along $AB$ (rightward):

   • $x$-component: $+2P$
   • $y$-component: $0$

3. Force $3P$ along $BC$ (downward):

   • $x$-component: $0$
   • $y$-component: $-3P$

4. Force $4P$ along $CD$ (leftward):

   • $x$-component: $-4P$
   • $y$-component: $0$

Step 3: Calculate the resultant force components

Sum the $x$- and $y$-components separately:

• Resultant $x$-component ($R_x$): $$R_x=0+2P+0-4P=-2P$$

• Resultant $y$-component ($R_y$): $$R_y=P+0-3P+0=-2P$$

Step 4: Find the magnitude of the resultant force

The magnitude of the resultant force $R$ is given by: $$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-2P{)}^2+(-2P{)}^2}=\sqrt{4P^2+4P^2}=\sqrt{8P^2}=2\sqrt{2}P$$

Step 5: Determine the direction of the resultant force

The direction of the resultant force is given by the angle $\theta$ it makes with the positive $x$-axis. This angle can be found using: $$\tan \theta =\frac{R_y}{R_x}=\frac{-2P}{-2P}=1$$ $$\theta =\arctan (1)=45^{\circ }$$

Since both $R_x$ and $R_y$ are negative, the resultant force acts in the third quadrant, making an angle of $225^{\circ }$ (or equivalently, $45^{\circ }$ below the negative $x$-axis).

Step 6: Find the line of action of the resultant force

To find the line of action of the resultant force, we calculate the moment of each force about a reference point (e.g., point $A$) and then find the perpendicular distance of the resultant from $A$.

Moments about point $A$:

1. Force $P$ along $DA$:

   • Acts through $A$, so its moment about $A$ is $0$.

2. Force $2P$ along $AB$:

   • Acts at a distance $a$ from $A$ (along $AB$).
   • Moment: $2P\times a$ (clockwise, negative by convention).

3. Force $3P$ along $BC$:

   • Acts at a distance $a$ from $A$ (along $BC$).
   • Moment: $-3P\times a$ (counterclockwise, positive by convention).

4. Force $4P$ along $CD$:

   • Acts at a distance $a$ from $A$ (along $CD$).
   • Moment: $4P\times a$ (clockwise, negative by convention).

Total moment about $A$: $$M_A=0-2Pa-3Pa+4Pa=-Pa$$

The resultant force $R=2\sqrt{2}P$ acts at a perpendicular distance $d$ from $A$ such that: $$R\times d=|M_A|$$ $$2\sqrt{2}P\times d=Pa$$ $$d=\frac{Pa}{2\sqrt{2}P}=\frac{a}{2\sqrt{2}}=\frac{a\sqrt{2}}{4}$$

The line of action of the resultant force is parallel to the diagonal $AC$ of the square but shifted by $\frac{a\sqrt{2}}{4}$ from $A$ toward the side $BC$.

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Final Answer

• Magnitude of the resultant force: $\boxed{2\sqrt{2}P}$
• Direction of the resultant force: $225^{\circ }$ (or $45^{\circ }$ below the negative $x$-axis).
• Line of action: The resultant force acts along a line parallel to the diagonal $AC$ of the square, at a perpendicular distance of $\frac{a\sqrt{2}}{4}$ from point $A$.