Chapter 4, Problem 27P

(a)

To determine

The net force on the object and its acceleration when two forces $F_1$ and $F_2$ act on an 18.5 kg object on a frictionless table top.

Answer to Problem 27P

Solution:

Net force is 18.97 N. Acceleration is $1.025{\text{ m/s}}^{\text{2}}$ .

Explanation of Solution

Given: An 18.5 kg object on a frictionless tabletop.

  $F_1=10.2\text{ N}$

  $F_2=16.0\text{ N}$

Formula used:

Force is a vector quantity. The resultant force is obtained by vector addition of the two forces.

  $F_{net}=\sqrt{F_1{}^2+F_2{}^2+2F_1{F}_2\text{cos}\theta }$

Calculation:

The angle between the given forces is $\theta =90°$ .

  $\text{cos}90°=0$

  $F_{Net}=\sqrt{F_1{}^2+F_2{}^2}$

  $=\sqrt{{(10.2)}^2+{\text{ (16)}}^2}$

Net force ( F_Net )=18.97 N. So, the net force is 18.97 N.

Now, to calculate its acceleration, use the given mass ( m ) which is equal to 18.5 kg.

  $m{a}_{net}=18.97$

  $a_{net}=\frac{18.97\text{ N}}{18.5\text{kg}}$

  $a_{net}=1.025{\text{ m/s}}^{\text{2}}$

Conclusion:

The net force on the object is 18.97 N and the acceleration is $1.025{\text{ m/s}}^{\text{2}}\text{.}$

(b)

To determine

The net force on the object and its acceleration when two forces act on an 18.5 kg object on a frictionless table top.

Answer to Problem 27P

Solution:

Net force is 14.82 N.

Acceleration is $0.75{\text{ m/s}}^{\text{2}}$ .

Explanation of Solution

Given: An 18.5 kg object on a frictionless tabletop.

  $F_1=10.2\text{ N}$

  $F_2=16.0\text{ N}$

Formula used:

Force is a vector quantity. The resultant force is obtained by vector addition of the two forces.

  $F_{net}=\sqrt{F_1{}^2+F_2{}^2+2F_1{F}_2\text{cos}\theta }$

Calculation:

The angle between the given forces is $\theta =120°$ .

  $\text{cos}120°=-\frac{1}{2}$

  $F_{ne}{}_t=\sqrt{F_1{}^2+F_2{}^2+2F_1{F}_2\cos {120}^{\text{o}}}$

  $\begin{array}{l}{F}_{net}=\sqrt{{(10.2\text{)}}^2+{\text{ (16)}}^2+2(10.2)(16.1)(-1/2)}\\ F_{net}=14.82\text{ N}\end{array}$

  $\begin{array}{l}m{a}_{net}=14.82\text{N}\\ a_{net}=\frac{14.82\text{N}}{18.5\text{kg}}\\ a_{net}=0.75{\text{m/s}}^{\text{2}}\end{array}$ .

Conclusion: The net force on the object is 18.97 N and the acceleration is $1.025{\text{ m/s}}^{\text{2}}\text{.}$