## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 2 - Section 2.4 - Inertia and Equilibrium: Newton's First Law of Motion - Practice Problem - Page 42: 2.8

#### Answer

The contact force due to the floor has a magnitude of $758~N$ and this force is directed at an angle of $81.7^{\circ}$ above the horizontal.

#### Work Step by Step

The horizontal component of the contact force due to the floor must be equal in magnitude to the force that is applied horizontally. Therefore, the horizontal component of the contact force due to the floor has a magnitude of $110~N$. That is: $F_x = 110~N$ The vertical component of the contact force due to the floor must be equal in magnitude to the weight. Therefore, the vertical component of the contact force due to the floor has a magnitude of $750~N$. That is: $F_y = 750~N$ and this force is directed upward. We can find the magnitude of the sum of the forces: $F = \sqrt{F_x^2+F_y^2} = \sqrt{(110~N)^2+(750~N)^2} = 758~N$ We can find the angle $\theta$ above the horizontal: $tan~\theta = \frac{750~N}{110~N}$ $\theta = tan^{-1}(\frac{750~N}{110~N})$ $\theta = 81.7^{\circ}$ The contact force due to the floor has a magnitude of $758~N$ and this force is directed at an angle of $81.7^{\circ}$ above the horizontal.

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