Question

The two forces${\overrightarrow{\mathbf{F}}}_1$and${\overrightarrow{\mathbf{F}}}_2$, as shown in Fig. 4–52 (a) and (b), (looking down) act on an 18.5 kg object on a frictionless tabletop. If${\mathbf{F}}_{\mathbf{1}}=\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{N}$,and${\mathbf{F}}_{\mathbf{2}}=\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{N}$, find the net force on the object and its acceleration for (a) and (b).

FIGURE 4-52 Problem 28.

Short answer

The net force and the acceleration are

(a) $19\mathrm{N}$and $1.03\mathrm{m}{/\mathrm{s}}^2$.

(b) $14.03\mathrm{N}$and$0.76\mathrm{m}{/\mathrm{s}}^2$.

Step-by-step solution

Step 1. Newton’s second law

According to Newton’s second law, “the force acting on a body is equal to the product of the body’s acceleration and mass”.Mathematically, it is given as

$F=ma$.

Here, $m$is the mass of the body, and $a$is the acceleration of the body.

Step 2. Given information

Given data:

The first force is $F_1=10.2\mathrm{N}$.

The second force is $F_2=16.0\mathrm{N}$.

The mass of the object is $m=18.5\mathrm{kg}$.

Step 3. Calculate the net force and acceleration for part (a)

(a)

The free-body diagram can be drawn as follows:

The net force on the object can be calculated as

$\begin{array}{c}{F}_{net}=\sqrt{{\left[F_1\cos \left(180°\right)\right]}^2+{\left[F_2\cos \left(180°\right)\right]}^2}\\ F_{net}=\sqrt{{\left[\left(10.2\mathrm{N}\right)\cos \left(180°\right)\right]}^2+{\left[\left(16.0\mathrm{N}\right)\cos \left(180°\right)\right]}^2}\\ F_{net}=\sqrt{360.04\mathrm{N}}\\ F_{net}=18.97\mathrm{N}\approx 19\mathrm{N}\end{array}$

The acceleration of the object can be calculated as

$\begin{array}{c}{F}_{net}=ma\\ \left(18.97\mathrm{N}\right)=\left(18.5\mathrm{kg}\right)a\\ a=1.03\mathrm{m}{/\mathrm{s}}^2\end{array}$

Thus, the net force and the acceleration of the object are $19\mathrm{N}$and$1.03\mathrm{m}{/\mathrm{s}}^2,$respectively.

Calculate the net force and acceleration for part (b)

(b)

The free-body diagram can be drawn as follows:

The net force on the object can be calculated as

$\begin{array}{l}{F}_{net}=\sqrt{{\left[F_1\cos 30°\right]}^2+{\left[F_2-F_1\sin 30°\right]}^2}\\ F_{net}=\sqrt{{\left[\left(10.2\mathrm{N}\right)\cos 30°\right]}^2+{\left[\left(16\mathrm{N}\right)-\left(10.2\mathrm{N}\right)\sin 30°\right]}^2}\\ F_{net}=14.03\mathrm{N}\end{array}$

The acceleration of the object can be calculated as

$\begin{array}{c}{F}_{net}=ma\\ \left(14.03\mathrm{N}\right)=\left(18.5\mathrm{kg}\right)a\\ a=0.76\mathrm{m}{/\mathrm{s}}^2\end{array}$

Thus, the net force and the acceleration of the object are $14.03\mathrm{N}$and$0.76\mathrm{m}{/\mathrm{s}}^2$, respectively.