Chapter 4, Problem 51A

(a)

To determine

Maximum mass that a winch can lift.

Answer to Problem 51A

Maximum mass a winch can lift is 204.08 kg.

Explanation of Solution

Given:

The given figure is shown below.

  

  $F_{max} =2000\text{ N}$

Formula used:

According to Newton’s 2^nd law, force is given by the formula,

  $F=ma$

Where,

m is mass of the body

a is acceleration of body

Calculation:

Maximum mass that can be lifted by winch is,

  $\begin{array}{l}{m}_{\max }=\frac{F_{\max }}{a}\\ =\frac{2000}{9.8}\left(\because a=\text{acceleration due to gravity}=9.8{\text{ m/s}}^{\text{2}}\right)\\ m_{\max }=204.08\text{kg}\end{array}$

(b)

To determine

The rescue’s and victim’s acceleration.

To draw: A free body diagram of the people being lifted.

Answer to Problem 51A

The acceleration of victim is 9.8 m/s² and acceleration of rescue is 0.2 m/s²_.

A free body diagram of the people being lifted is shown in Figure 1.

Explanation of Solution

Given:

The given figure is shown below.

  

  $F_{winch}=1200\text{N}$

Calculation:

From part (a), the maximum mass is

  $m_{\max }\text{=}\text{204}\text{.08}\text{kg}$

Winch force is acting in the upward direction that is along positive y -direction

Mass of victims = 120 kg

Force exerted by the victim is,

  $\begin{array}{c}{F}_{victim}=m{a}_v\\ =\left(120\right)\left(9.8\right)\\ =1176\text{N}\end{array}$

The direction of force in the downward direction that is along negative y -direction.

Total force exerted on victim is,

  $\begin{array}{c}{F}_{total}=F_{winch}+F_{victim}\\ =(1200\text{N})+(-1176\text{N})\\ =24\text{N}\end{array}$

(Considering force along positive y -direction is positive and along negative y -direction is negative)

Positive sign of the force signifies that the resultant force in the direction of positive y- direction.

Acceleration of rescuer is

  $\begin{array}{l}{a}_r=\frac{F_{total}}{m}=\frac{24}{120}\\ a_r=0.2{\text{m/s}}^{\text{2}}\\ \text{And}\text{acceleration}\text{of}\text{victim}\text{is}\\ a_v=9.8\text{}{\text{m/s}}^{\text{2}}\end{array}$

The free-body diagram is shown below.

  

Figure 2

(c)

To determine

Time taken to pull the people up to the helicopter.

Answer to Problem 51A

Time taken to pull the people up to the helicopter is 10.95 sec.

Explanation of Solution

Given:

The given figure is shown below.

  

Initially victims are stationary

  $u=0\text{ m/s}$

Distance between rescue and victim is 12 m.

  $s=12\text{ m}$

Rescue’s acceleration is,

  $a_r=0.2{\text{ m/s}}^{\text{2}}$

Formula used:

The equation of motion is,

  $s=ut+\frac{1}{2}a{t}^2$

Calculation:

Substituting the values in above equation.

  $\begin{array}{l}12=(0)t+\frac{1}{2}(0.2)t^2\\ t^2=\frac{(12)(2)}{0.2}\\ \therefore t=10.95\sec \end{array}$