5.5 Newton’s third law  (Page 4/6)

a. action: Earth pulls on the Moon, reaction: Moon pulls on Earth; b. action: foot applies force to ball, reaction: ball applies force to foot; c. action: rocket pushes on gas, reaction: gas pushes back on rocket; d. action: car tires push backward on road, reaction: road pushes forward on tires; e. action: jumper pushes down on ground, reaction: ground pushes up on jumper; f. action: gun pushes forward on bullet, reaction: bullet pushes backward on gun.

a. The rifle (the shell supported by the rifle) exerts a force to expel the bullet; the reaction to this force is the force that the bullet exerts on the rifle (shell) in opposite direction. b. In a recoilless rifle, the shell is not secured in the rifle; hence, as the bullet is pushed to move forward, the shell is pushed to eject from the opposite end of the barrel. c. It is not safe to stand behind a recoilless rifle.

a. $F_{\text{net}}=2.64\times {10}^7\text{N;}$ b. The force exerted on the ship is also $2.64\times {10}^7\text{N}$ because it is opposite the shell’s direction of motion.

Because the weight of the history book is the force exerted by Earth on the history book, we represent it as ${\overrightarrow{F}}_{\text{EH}}=\mathrm{-14}\widehat{j}\text{N}\text{.}$ Aside from this, the history book interacts only with the physics book. Because the acceleration of the history book is zero, the net force on it is zero by Newton’s second law: ${\overrightarrow{F}}_{\text{PH}}+{\overrightarrow{F}}_{\text{EH}}=\overrightarrow{0},$ where ${\overrightarrow{F}}_{\text{PH}}$ is the force exerted by the physics book on the history book. Thus, ${\overrightarrow{F}}_{\text{PH}}=\text{−}{\overrightarrow{F}}_{\text{EH}}=\text{−}\left(\mathrm{-14}\widehat{j}\right)\text{N}=14\widehat{j}\text{N}\text{.}$ We find that the physics book exerts an upward force of magnitude 14 N on the history book. The physics book has three forces exerted on it: ${\overrightarrow{F}}_{\text{EP}}$ due to Earth, ${\overrightarrow{F}}_{\text{HP}}$ due to the history book, and ${\overrightarrow{F}}_{\text{DP}}$ due to the desktop. Since the physics book weighs 18 N, ${\overrightarrow{F}}_{\text{EP}}=\mathrm{-18}\widehat{j}\text{N}\text{.}$ From Newton’s third law, ${\overrightarrow{F}}_{\text{HP}}=\text{−}{\overrightarrow{F}}_{\text{PH}},$ so ${\overrightarrow{F}}_{\text{HP}}=\mathrm{-14}\widehat{j}\text{N}\text{.}$ Newton’s second law applied to the physics book gives $\sum \overrightarrow{F}=\overrightarrow{0},$ or ${\overrightarrow{F}}_{\text{DP}}+{\overrightarrow{F}}_{\text{EP}}+{\overrightarrow{F}}_{\text{HP}}=\overrightarrow{0},$ so ${\overrightarrow{F}}_{\text{DP}}=\text{−}\left(\mathrm{-18}\widehat{j}\right)-\left(\mathrm{-14}\widehat{j}\right)=32\widehat{j}\text{N}\text{.}$ The desk exerts an upward force of 32 N on the physics book. To arrive at this solution, we apply Newton’s second law twice and Newton’s third law once.