Question

In Figure P3.55, a spider is resting after starting to spin its web. The gravitational force on the spider makes it exert a downward force of $\mathbf{0}.\mathbf{150}\mathbf{N}$on the junction of the three strands of silk. The junction is supported by different tension forces in the two strands above it so that the resultant force on the junction is zero. The two sloping strands are perpendicular, and we have chosen the xand ydirections to be along them. The tension$T_x$is$0.127N$. Find

(a) the tension$T_y$,

(b) the angle the x axis makes with the horizontal, and

(c) the angle the y axis makes with the horizontal.

Short answer

(a)The tension$T_y$is$0.80\text{N}$

(b)The angle of the x axis with horizontal is ${\theta }_x=57.9^{°}$

(c)The angle of the y axis with horizontal is ${\theta }_y=32.1^{°}.$

Step-by-step solution

Definition of a force vector and expression used in solution

• A force vector is a representation of both magnitude and direction of a force. This is in contrast to merely stating the force's magnitude, which is referred to as a scalar number.
• A vector is often depicted by an arrow pointing in the force's direction and having a length proportionate to the magnitude of the force.
• _{${\overrightarrow{\mathbf{F}}}_{net}=\overrightarrow{\mathbf{W}}+{\overrightarrow{\mathbf{T}}}_x+{\overrightarrow{\mathbf{T}}}_y$}
  

Determine the tension Ty

(a)

First, write all the forces in unit- vector form:

$$\begin{gathered} {\overrightarrow{\mathbf{T}}}_x=T_x\hat{\mathbf{i}}=0.127\hat{\mathbf{i}}\text{N} \\ {\overrightarrow{\mathbf{T}}}_y=T_y\hat{\mathbf{j}}\text{N} \\ \overrightarrow{\mathbf{W}}=-w\cos \theta \hat{\mathbf{i}}-w\sin \theta \hat{\mathbf{j}} \\ \overrightarrow{\mathbf{W}}=-0.150\cos \theta \hat{\mathbf{i}}-0.150\sin \theta \hat{\mathbf{j}}\text{N} \end{gathered}$$

Substitute the known values in force

Now, get the resultant force on the junction, which is the summation of all forces acting on it ,

$$\begin{gathered} {\overrightarrow{\mathbf{F}}}_{\text{net}}=\overrightarrow{\mathbf{W}}+{\overrightarrow{\mathbf{T}}}_x+{\overrightarrow{\mathbf{T}}}_y \\ =(-0.150\cos \theta \hat{\mathbf{i}}-0.150\sin \theta \hat{\mathbf{j}})+(0.127\hat{\mathbf{i}})+\left(T_y\hat{\mathbf{j}}\right) \\ =(-0.150\cos \theta +0.127)\hat{\mathbf{i}}+\left(-0.150\sin \theta +T_y\right)\hat{\mathbf{j}} \end{gathered}$$

But, the resultant force ${\overrightarrow{\mathbf{F}}}_{\text{net}}$is zero, so :

$$\begin{gathered} {\left(F_{net}\right)}_x=0 \\ {\left(F_{net}\right)}_y=0 \end{gathered}$$

Now,thetwo equations are:

Solving equation$\left(1\right)$for $\theta$,

$$\begin{gathered} \theta ={\cos }^{-1}\frac{0.127}{0.150} \\ \theta =32.1° \end{gathered}$$

Substituting the value of $\theta$into equation $\left(2\right)$and solving for$T_y$, $$\begin{gathered} T_y=0.150\sin 32.1=0.080\text{N} \\ \therefore T_y=0.80\text{N} \end{gathered}$$

Hence, the tension $T_y$is $0.80\text{N}$.

Determine the angle of the  axis makes with the horizontal

(b)

As from the figure above, the angle the $X-$axis makes with the horizontal is:

$$\begin{gathered} {\theta }_x=90-\theta \\ {\theta }_x=90-32.1 \\ {\theta }_x=57.9^{°} \end{gathered}$$

Therefore, the angle of the x axis with horizontal is${\theta }_x=57.9^{°}$

Find the angle of the y axis makes with the horizontal

(c)

The angle the$y-$axis makes with the horizontal is:

$$\begin{gathered} {\theta }_y=\theta =32.1^{°} \\ \therefore {\theta }_y=32.1^{°} \end{gathered}$$

Therefore, the angle of the y axis with horizontal is ${\theta }_y=32.1^{°}.$