Chapter 5.1, Problem 10PP

To determine

The x- and y- components of the initial net force exerted by A and C on the box.

Answer to Problem 10PP

  $F_{\text{net on the box, x}}=0.0\text{ N}$

  $F_{\text{net on the box, y}}=0.8\text{ N}$

Explanation of Solution

Given:

Net force exerted by A on the box is, $F_{A\text{ on box}}=20.4\text{ N}$ at an angle $\text{120}{\text{.0}}^0$ to the positive x-direction.

Net force exerted by C on the box is, $F_{\text{C on box}}=17.7\text{ N}$ at an angle $\text{55}{\text{.0}}^0$ to the positive x-direction.

Mass of the box is $m=3.20\text{ kg}$ .

Calculation:

x-component of force exerted by A on the box is,

  $F_{A\text{ on the box, x}}=F_{A\text{ on box}}\cos {\theta }_A$

  $=\left(20.4\text{ N}\right)\left(\cos {120.0}^0\right)$

  $=\left(20.4\text{ N}\right)\left(-0.5\right)$

  $=-10.2\text{ N}$

y-component of force exerted by A on the box is,

  $F_{A\text{ on the box, y}}=F_{A\text{ on box}}\sin {\theta }_A$

  $=\left(20.4\text{ N}\right)\left(\sin {120.0}^0\right)$

  $=\left(20.4\text{ N}\right)\left(0.86\right)$

  $=17.7\text{ N}$

x-component of force exerted by C on the box is,

  $F_{\text{C on the box, x}}=F_{\text{C on box}}\cos {\theta }_C$

  $=\left(17.7\text{ N}\right)\left(\cos {55.0}^0\right)$

  $=\left(17.7\text{ N}\right)\left(0.57\right)$

  $=10.2\text{ N}$

y-component of force exerted by C on the box is,

  $F_{\text{C on the box, y}}=F_{\text{C on box}}\sin {\theta }_C$

  $=\left(\text{17}\text{.7 N}\right)\left(\sin {55.0}^0\right)$

  $=\left(17.7\text{ N}\right)\left(0.81\right)$

  $=14.5\text{ N}$

Force on the box in the positive x-direction is

  ${\text{F}}_{g,x}=0\text{ N}$

Force on the box in the positive y-direction is,

  ${\text{F}}_{g,y}=-mg$

  $=\left(-3.20\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)$

  $=-31.4\text{ N}$

Now, the x-component of initial net force exerted by A and C on the box is

  $F_{\text{net on the box, x}}=F_{A\text{ on the box, x}}+F_{\text{C on the box, x}}+{\text{F}}_{g,x}$

  $=-10.2\text{ N+10}\text{.2 N+0}\text{.0 N}$

  $\text{=0}\text{.0 N}$

Now, the y-component of initial net force exerted by A and C on the box is

  $F_{\text{net on the box, y}}=F_{A\text{ on the box, y}}+F_{\text{C on the box, y}}+{\text{F}}_{g,y}$

  $=17.7\text{ N+14}\text{.5 N-31}\text{.4 N}$

  $\text{=0}\text{.8 N}$

Conclusion:

The x-component of initial net force exerted by A and C on the box is $\text{0}\text{.0 N}$ .

The y-component of initial net force exerted by A and C on the box is $\text{0}\text{.8 N}$ .