Question

you push a book across a table to your friend with a force of 6.5 n e. the force of friction on the book is 4.5 n w. the normal force and the force of gravity have a magnitude of 7.5 n. draw the fbd of the book and calculate the net force on the book.

Explanation:

Step1: Analyze horizontal forces

The applied force is \( F_{app} = 6.5\,\text{N [E]} \) and friction is \( F_{fric} = 4.5\,\text{N [W]} \). Since east and west are opposite directions, we subtract their magnitudes for net horizontal force: \( F_{net,x} = F_{app} - F_{fric} = 6.5 - 4.5 = 2.0\,\text{N [E]} \).

Step2: Analyze vertical forces

Normal force \( F_N = 7.5\,\text{N [up]} \) and gravity \( F_g = 7.5\,\text{N [down]} \). These are opposite, so net vertical force: \( F_{net,y} = F_N - F_g = 7.5 - 7.5 = 0\,\text{N} \).

Step3: Find total net force

Since vertical net force is 0, total net force is equal to horizontal net force: \( F_{net} = \sqrt{F_{net,x}^2 + F_{net,y}^2} = \sqrt{2.0^2 + 0^2} = 2.0\,\text{N [E]} \).

(For FBD: Draw a dot (book). Arrows: right for applied force (6.5 N), left for friction (4.5 N), up for normal (7.5 N), down for gravity (7.5 N).)

Answer:

The net force on the book is \( \boldsymbol{2.0\,\text{N [East]}} \).