Question

Figure 5-62 is an overhead view of a$\mathbf{12}\mathbf{}\mathit{k}\mathit{g}$12tire that is to be pulled by three horizontal ropes. One rope’s force $\mathbf{(}{\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathbf{50}\mathbf{}\mathit{N}\mathbf{)}$is indicated. The forces from the other ropes are to be oriented such that the tire’s acceleration magnitude ais

least. What is that least if: (a)${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{}\mathit{N}$${\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{20}\mathbf{}\mathit{N}$; (b) ${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{}\mathit{N}\mathbf{,}\mathbf{}{\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{10}\mathbf{}\mathit{N}$; and

(c) ${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{30}\mathbf{}\mathit{N}$?

Short answer

1. $a=0m/s^2$
   
2. $a=0.83m/s^2$
   
3. $a=0m/s^2$
   

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} M=12kg \\ {F}_1=50N \end{gathered}$$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.Acceleration can be calculated by using Newton’s second law.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\underset{}{\sum Ma}$

where, $F$ is the net force, $M$is mass and $a$ is an acceleration.

(a) Determining the acceleration when F2=30 N, F3=20 N

When$F_2=30Nand{F}_3=20N$

$a=\frac{Netforce}{mass}$

$a=\frac{F_1-F_2-F_3}{m}$

$a=\frac{50-30-20}{12}$

$a=0m/s^2$

Hence, the least acceleration for is$F_2=30Nand{F}_3=20N$is a$a=0m/s^2$

(b) Determining the acceleration when F2=30 N, F3=10 N

When $F_2=30Nand{F}_3=10N$,

$a=\frac{Netforce}{mass}$

$a=\frac{F_1-F_2-F3}{m}$

$a=\frac{50-30-10}{12}$

$a=0.83m/s^2$

Hence, the least acceleration for is$F_2=30Nand{F}_3=10N$is a$a=0.83m/s^2$

(c) Determining for F2=F3=30 N

When, $F_2=F_3=30N$

Now, apply $F_{2}and{F}_3$in the opposite direction making an angle$\theta$with$x-axis$

For least acceleration:

$$\begin{gathered} \sum _{}{F}_x=0 \\ 50-30\times \cos \left(\theta \right)-30\times \cos \left(\theta \right)=0 \\ 50=60\cos \left(\theta \right) \\ \theta =34° \end{gathered}$$

Forces $\overrightarrow{F_2}$and $\overrightarrow{F_3}$should be arranged with an angle $34°$with the negative horizontal axis to produce least acceleration. For $\theta =34°$least acceleration will be$a=0m/s^2$.