Question

The only two forces acting on a body have magnitudes of $\mathbf{20}\mathbf{}\mathit{N}\mathbf{}$and $\mathbf{35}\mathbf{}\mathit{N}$and directions that differ by ${\mathbf{80}}^{\mathbf{\circ }}$. The resulting acceleration has a magnitude of role="math" localid="1655525874259" $\mathbf{20}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What is the mass of the body?

Short answer

Mass of the body is$m=2.6kg.$

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} F_1=20N \\ {F}_2=35N \end{gathered}$$

The angle between $F_1$and $F_2$is$\theta ={80}^{\circ }$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.Also this involves resolution of vectors.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\sum _{}^{}Ma\left(i\right)$

Determining the mass of the body

Free Body Diagram:

$$\begin{gathered} \overrightarrow{F_1}=20\hat{i} \\ \overrightarrow{F_2}=35\cos \left(80\right)\hat{i}+35\sin \left(80\right)\hat{j} \\ \overrightarrow{F_2}=6.1\hat{i}+34.5\hat{j} \end{gathered}$$

Now resultant force will be:

role="math" localid="1655527433958" $$\begin{gathered} \overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2} \\ \overrightarrow{F}=(20+6.1)\hat{i}+34.5\hat{j} \\ \overrightarrow{F}=26.1\hat{i}+34.5\hat{j} \end{gathered}$$

Magnitude of resultant is,

$$\begin{gathered} \left|\stackrel{\rightharpoonup }{F}\right|=\sqrt{{\left(26.1\right)}^2+{\left(34.5\right)}^2} \\ =43.3N \end{gathered}$$

Thus, using equation (i),

$$\begin{gathered} 43.3=m\times 20 \\ m=2.16kg \end{gathered}$$

Hence, the mass of the body is$m=2.16kg$