Chapter 4, Problem 15P

(a)

To determine

The net force on the object

Answer to Problem 15P

The net force on the object is $30\text{N}$ to the right.

Explanation of Solution

Write an expression for the net force.

$F_a=F_1+F_2$ (I)

Here, $F_a$ is the net force on the object, $F_1$ is the first force acting to the right and $F_2$ is the second force acting to the left.

Conclusion:

Substitute $40\text{N}$ for $F_1$, and $-10\text{N}$ for $F_2$ in equation (I) to find $F_a$.

$\begin{array}{c}{F}_a=40\text{N}+-\left(10\text{N}\right)\\ =40\text{N}-10\text{N}\\ =3\text{0}\text{N}\end{array}$

Thus, the net force on the object is $30\text{N}$ to the right.

(b)

To determine

The net force on the object

Answer to Problem 15P

The net force on the object is $0\text{N}$.

Explanation of Solution

Write an expression for the net force.

$\left|F_b\right|=\sqrt{{\left(F_1+F_2\right)}^2+{\left(F_3+F_4\right)}^2}$ (II)

Here, $F_b$ is the net force on the object, $F_1$ is the first force acting to the right, $F_2$ is the second force acting to the left, $F_3$ is the third force acting upwards and $F_4$ is the fourth force acting downwards.

Conclusion:

Substitute $10\text{N}$ for $F_1$, $-10\text{N}$ for $F_2$, $18\text{N}$ for $F_3$, and $-18\text{N}$ for $F_4$ in equation (II) to find $F_b$.

$\begin{array}{c}\left|F_b\right|=\sqrt{{\left(10\text{N}+\left(-10\text{N}\right)\right)}^2+{\left(18\text{N+}\left(-18\text{N}\right)\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(0\right)}^2}\text{N}\\ =0\text{N}\end{array}$

Thus, the net force on the object is $0\text{N}$.

(c)

To determine

The net force on the object

Answer to Problem 15P

The net force on the object is $18\text{N}$ downwards.

Explanation of Solution

Write an expression for the net force.

$\left|F_c\right|=\sqrt{{\left(F_1+F_2\right)}^2+{\left(F_3\right)}^2}$ (III)

Here, $F_c$ is the net force on the object, $F_1$ is the first force acting to the right, $F_2$ is the second force acting to the left and $F_3$ is the third force acting downwards.

Conclusion:

Substitute $10\text{N}$ for $F_1$, $-10\text{N}$ for $F_2$, and $18\text{N}$ for $F_3$ in equation (III) to find $F_c$.

$\begin{array}{c}\left|F_c\right|=\sqrt{{\left(10\text{N}+\left(-10\text{N}\right)\right)}^2+{\left(-18\text{N}\right)}^2}\\ =\sqrt{{18}^2}\text{N}\\ =18\text{N}\end{array}$

Thus, the net force on the object is $18\text{N}$ downwards.