Newton's Third Law — Question 1

Section 5-7 · Physics pages 732–733

Question 1 The reaction to weight acts on Earth A book rests on a table. Earth pulls the book downward with 10 N; the table pushes it upward with 10 N. book Earth table 10 N 10 N 10 N 10 N Interaction map: arrows indicate agent to recipient, not physical force directions. (a) Identify the third-law partner of Earth’s force on the book. (b) Identify the partner of the table’s force on the book. (c) Why are the two stated forces on the book not a third-law pair?
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Question 1 - Solution The reaction to weight acts on Earth book Earth table 10 N 10 N 10 N 10 N (a) Reverse agent and recipient for gravity.⃗ FB→E = −⃗FE→B ⇒⃗ FB→E = +10ˆy N The book pulls Earth upward toward the book with 10 N. The two gravitational forces belong to the same book-Earth interaction. (b) Reverse the contact interaction.⃗ FB→T = −⃗FT→B = −10ˆy N The book pushes the table downward with 10 N. That is the partner of the table’s upward force on the book. (c) Compare the bodies acted upon.⃗ FE→B +⃗FT→B =⃗0 (both recipients are B) The two given forces both act on the book and come from different agents. They balance here because the book has zero acceleration. A third-law pair always acts on different recipients and exists even when either recipient accelerates.

Question 1

The reaction to weight acts on Earth

A book rests on a table. Earth pulls the book downward with 10 N; the table pushes it upward with 10 N.

Diagram shown in the original SVG worksheet.

  1. Identify the third-law partner of Earth’s force on the book.

  2. Identify the partner of the table’s force on the book.

  3. Why are the two stated forces on the book not a third-law pair?

Question 1 - Solution

The reaction to weight acts on Earth

Diagram shown in the original SVG worksheet.

(a) Reverse agent and recipient for gravity.

F→B→E=−F→E→B⇒F→B→E=+10ŷN\vec F_{B\to E}=-\vec F_{E\to B}\ \Rightarrow\ \boxed{\vec F_{B\to E}=+10\hat y\,\mathrm N} The book pulls Earth upward toward the book with 10 N. The two gravitational forces belong to the same book-Earth interaction.

(b) Reverse the contact interaction.

F→B→T=−F→T→B=−10ŷN\vec F_{B\to T}=-\vec F_{T\to B}=-10\hat y\,\mathrm N The book pushes the table downward with 10 N. That is the partner of the table’s upward force on the book.

(c) Compare the bodies acted upon.

F→E→B+F→T→B=0→(both recipients are B)\vec F_{E\to B}+\vec F_{T\to B}=\vec0\quad\text{(both recipients are B)} The two given forces both act on the book and come from different agents. They balance here because the book has zero acceleration. A third-law pair always acts on different recipients and exists even when either recipient accelerates.