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. k/u c ans: 2.0 n e

Explanation:

Step1: Analyze vertical forces

The normal force \(F_N = 7.5\ N\) (upward) and the force of gravity \(F_g=7.5\ N\) (downward). Since \(F_N\) and \(F_g\) are equal in magnitude and opposite in direction, their resultant in the vertical direction is \(F_{vertical}=F_N - F_g=7.5 - 7.5=0\ N\).

Step2: Analyze horizontal forces

The applied force \(F_{applied}=6.5\ N\) [E] and the frictional force \(F_f = 4.5\ N\) [W]. Taking east as positive, the net force in the horizontal direction is \(F_{horizontal}=F_{applied}-F_f\). Substitute the values: \(F_{horizontal}=6.5 - 4.5\).

Answer:

The net force on the book is \(2.0\ N\) [E].