Question

Question: Forces${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{\text{,}}{\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{}\mathbf{\text{and}}{\overrightarrow{\mathbf{F}}}_{\mathbf{3}}$ act on the structure of Fig. 12-33, shown in an overhead view. We wish to put the structure in equilibrium by applying a fourth force, at a point such as P. The fourth force has vector components${\overrightarrow{\mathbf{F}}}_h\mathbf{}\mathbf{\text{and}}{\overrightarrow{\mathbf{F}}}_v$ . We are given that a = 2.0 m,b = 3.0m , c = 1 0 m , ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{\text{N,}}{\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\text{Nand}}{\overrightarrow{\mathbf{F}}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{N}}$Find (a) F_h , (b) F_v, and (c) d.

Short answer

Answer

$$\begin{gathered} a.F_h=5.0\text{N} \\ b.F_v=30\text{N} \\ c.d=1.3m \end{gathered}$$

Step-by-step solution

Understanding the given information  

$$\begin{gathered} \text{a}=2.0\text{m} \\ \text{b}=3.0\text{m} \\ \text{c}=1.0\text{m} \\ {\text{F}}_1=20\text{N} \\ {\text{F}}_2=10\text{N} \\ {F}_3=5.0\text{N} \end{gathered}$$

Concept and formula used in the given question

By applying the equations of static equilibrium, you can get the equations in terms of unknown forces. By solving these equations, you can find the values of unknown forces and distance. The equations used are given below.

Static Equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

(a) Calculation for the Fh

Using the given figure in the problem and applying static equilibrium conditions:
$$\begin{gathered} \sum {\text{F}}_{\text{x}}=0 \\ {\text{F}}_{\text{h}}-{\text{F}}_3=0······\left(1\right) \\ \sum {\text{F}}_{\text{y}}=0 \\ {\text{F}}_{\text{v}}-{\text{F}}_1-{\text{F}}_2=0······\left(2\right) \\ \sum =0 \\ \left({\text{F}}_{\text{v}}\times \text{d}\right)-\left({\text{F}}_2\times \text{b}\right)-\left({\text{F}}_3\times \text{a}\right)=0······\left(3\right) \end{gathered}$$

From equation (1):

$F_h-5=0$

Hence, $F_h-5=0$

(b) Calculation for the  Fv

From equation (2):

$F_v-20-10=0$

Hence, $F_v=30N$

(c) Calculation for the  d

From equation (3):

$$\begin{gathered} \left(30\times d\right)-\left(10\times 3\right)-\left(5\times 2\right)=0 \\ d=1.3\text{m} \end{gathered}$$

Hence, d = 1.3 m