Chapter 2.1, Problem 2.3P

(a)

To determine

The magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket using the parallelogram law by graphical method.

Answer to Problem 2.3P

The magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket is $\underset{\_}{20.1\text{kN}}$ and $\underset{\_}{21.2°}$ below the horizontal.

Explanation of Solution

Given information:

The tension (P) in the member B is 10 kN at an angle $\left(\theta \right)$ of $25°$ with the horizontal axis in the clockwise direction.

The tension (Q) in the member C is 15 kN at an angle $\left(\varphi \right)$ of $50°$ with the horizontal axis in the anticlockwise direction.

Calculation:

Calculate the angle $\left(\beta \right)$ between the tension for Q and the vertical surface:

$\beta =90°-\varphi$

Substitute $50°$ for $\varphi$.

$\begin{array}{c}\beta =90°-50°\\ =40°\end{array}$

Apply the parallelogram law for the force P and Q using the procedure as follows:

• Draw the tension force P (10 kN) using the horizontal angle $25°$ from the origin point O with the coordinates $\left(10\cos {25}^{°},10\sin {25}^{°}\right)$.
• Draw the tension force Q (15 kN) using the vertical angle $40°$ from the origin point O with the coordinates $\left(15\sin {40}^{°},-15\cos {40}^{°}\right)$.
• Draw a perpendicular line to the line OP from the force point P.
• Draw a perpendicular line to the line OQ from the force point Q and make it intersect with the perpendicular line OP.
• Draw a resultant force line by connecting the point O with the point S. Measure the angle $\alpha$ made by the horizontal axis with the resultant force line. It represents a parallelogram.
• Measure the magnitude of resultant and angle $\alpha$ using the geometry from the parallelogram.

Sketch the parallelogram with the force P and Q as in Figure (1).

From Figure (1),

Measure the magnitude and the direction of the resultant $\left(R\right)$.

Thus, the magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket using the parallelogram law is $\underset{\_}{20.1\text{kN}}$ and $\underset{\_}{21.2°}$.

(b)

To determine

The magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket using the triangle rule by graphical method.

Answer to Problem 2.3P

The magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket using the triangle law are $\underset{\_}{20.1\text{kN}}$ and $\underset{\_}{21.2°}$.

Explanation of Solution

Given information:

The tension (P) in the member B is 10 kN at an angle $\left(\theta \right)$ of $25°$ with the horizontal axis in the clockwise direction.

The tension (Q) in the member C is 15 kN at an angle $\left(\varphi \right)$ of $50°$ with the horizontal axis in the anticlockwise direction.

Calculation:

Apply the triangle law for the force P and Q using the procedure as follows:

• Draw the tension force P (10 kN) using the horizontal angle $25°$ from the origin point O with the co-ordinates $\left(10\cos {25}^{°},10\sin {25}^{°}\right)$.
• Draw the tension force Q (15 kN) using the horizontal angle $50°$ from the point P with the co-ordinates $\left(15\cos {50}^{°},-15\sin {50}^{°}\right)$.
• Draw a resultant force line by connecting the point O with the point Q. It is enclosed with triangle. The resultant force R makes an angle $\alpha$ with horizontal.
• Measure the magnitude of resultant and angle $\alpha$ using the geometry from the triangle.

Sketch the triangle with the force P and Q as in Figure (2).

From Figure (2),

Measure the magnitude and the direction of the resultant $\left(R\right)$ force.

Therefore, the magnitude $\left(R\right)$ and direction $\left(\alpha \right)$ of the resultant force exerted on the bracket using the triangle law is $\underset{\_}{20.1\text{kN}}$ and $\underset{\_}{21.2°}$.