Chapter 6, Problem 55QAP

To determine

The $F_{\max }$ in terms of m and v with the help of graph.

Answer to Problem 55QAP

  $F_{\max }=138.27\text{ N}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Final velocity = }v\\ \text{Mass of sled = }22.0\text{ kg}\end{array}$

Formula used:

  $\begin{array}{l}F=ma\\ F=\text{Force}\\ a=\text{acceleration}\\ m=\text{mass}\end{array}$

  $\begin{array}{l}{v}^2=u^2+2as\\ v=\text{Final velocity}\\ u=\text{initial velocity}\\ a=\text{acceleration}\\ s=\text{distance}\end{array}$

Calculation:

  $F_{\max }=3f$

  $\begin{array}{l}\text{From }x=0\text{ to }x=30\\ F=ma\\ F_{\max }=22.0\text{ kg*}a\\ a=\frac{F_{\max }}{22}\end{array}$

  $\begin{array}{l}{v}^2=u^2+2as\\ v_{30}=\text{ velocity at }x=30\\ v_{30}{}^2=0+2*\frac{F_{\max }}{22}*30\text{ m}\\ v_{30}=\sqrt{\frac{60F_{\max }}{22}}\end{array}$

  $\begin{array}{l}\text{From }x=30\text{ to }x=55\\ \text{Acceleration = }{a}_{30-55}\\ \text{Force = }{F}_{avg(30-55)}\\ \text{Consider,}\\ F_{\max }=3x\\ Force\text{ at x=55 = }{F}_{55}\text{ = -}x\end{array}$

  $\begin{array}{l}{\text{F}}_{\text{avg(30-55)}}=\frac{F_{55}-F_{\max }}{2}=-\frac{f-3f}{2}=-f\\ {\text{F}}_{\text{avg(30-55)}}\text{ in turms of }{F}_{\max },\\ F_{\max }=3f\\ \frac{F_{\max }}{3}=f\\ {\text{F}}_{\text{avg(30-55)}}=-x\\ {\text{F}}_{\text{avg(30-55)}}=-\frac{F_{\max }}{3}\\ F=ma\\ F_{avg(30-55)}=22.0kg*a_{avg(30-55)}\\ F_{avg(30-55)}=-\frac{F_{\max }}{3}\\ -\frac{F_{\max }}{3}=22.0kg*a_{avg(30-55)}\\ a=-\frac{F_{\max }}{22*3}\end{array}$

  $\begin{array}{l}For\text{ this point, u=}\sqrt{\frac{60F_{\max }}{22}}\\ v^2=u^2+2as\\ v_{55}=\text{ velocity at }x=55\\ v_{55}{}^2=\frac{60F_{\max }}{22}+2*(-\frac{F_{\max }}{22*3})*25\text{ m}\\ v_{55}{}^2=\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}\end{array}$

  $\begin{array}{l}\text{From }x=55\text{ to }x=90\\ F_{55-90}=\text{Force from x=55 to x=90=}-f\\ a_{55-90}=\text{acceleration from x=55 to x=90}\\ F_{55-90}\text{ in turms of }{F}_{\max },\\ \frac{F_{\max }}{3}=f\\ F_{55-90}=-\frac{F_{\max }}{3}\\ F=ma\\ -\frac{F_{\max }}{3}=22.0\text{ kg*}a\\ a_{55-90}=-\frac{F_{\max }}{22*3}\end{array}$

  $\begin{array}{l}For\text{ this point, }{u}^2\text{=}\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}\\ v^2=u^2+2as\\ v_{90}=\text{ velocity at }x=90\\ v_{55}{}^2=\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}+2*(-\frac{F_{\max }}{22*3})*35\text{ m}\\ v_{55}{}^2=\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}-\frac{70F_{\max }}{22*3}\end{array}$

  $\begin{array}{l}\text{From }x=90\text{ to }x=100\\ \text{Acceleration = }{a}_{90-100}\\ \text{Force = }{F}_{avg(90-100)}\\ F_{avg(90-100)}=\frac{0-(-f)}{2}=0.5f\\ \frac{F_{\max }}{3}=f\\ F_{avg(90-100)}=\frac{F_{\max }}{6}\\ F=ma\\ F_{avg(90-100)}=22.0\text{ kg*}{a}_{90-100}\\ F_{avg(90-100)}=\frac{F_{\max }}{6}\\ \frac{F_{\max }}{6}=22.0\text{ kg*}{a}_{90-100}\\ a=\frac{F_{\max }}{22*6}\end{array}$

  $\begin{array}{l}\text{For this point, }{u}^2\text{=}\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}-\frac{70F_{\max }}{22*3}\\ v^2=u^2+2as\\ v_{100}=\text{ velocity at }x=100=12.5{\text{ ms}}^{\text{-1}}\\ v_{100}{}^2=\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}-\frac{70F_{\max }}{22*3}+2*\frac{F_{\max }}{22*6}*10\text{ m}\\ v_{100}{}^2=\frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}-\frac{70F_{\max }}{22*3}+\frac{20F_{\max }}{22*6}={(}^{12.5}\\ \frac{60F_{\max }}{22}-\frac{50F_{\max }}{22*3}-\frac{70F_{\max }}{22*3}+\frac{20F_{\max }}{22*6}={(}^{12.5}\\ 156.25=2.73F_{\max }-0.75F_{\max }-1.00F_{\max }+0.15F_{\max }\\ 1.13F_{\max }=156.25\\ F_{\max }=138.27\text{ N}\end{array}$

Conclusion:

  $F_{\max }=138.27\text{ N}$