Question

An object of mass is suspended from the ceiling of an accelerating truck as shown in Figure P6.21. Taking a , find (a) the angle that the string makes with the vertical and (b) the tension T in the string.

Short answer

The solution is

(a)$\theta =17.0°$

(b) The string has a tension of$5.13N$

Step-by-step solution

Given data

The mass of the thing suspended from the car's ceiling$m=0.500kg$

Acceleration of the car$a=3.00{\text{m/s}}^2$

Concept Introduction

Any object remains in the state of motion or rest until and unless affected by an external nonequilibrium force.

Calculation of the angle

(a)

mass$m=0.500kg$

acceleration$a=3.00{\text{m/s}}^2$

by newton’s second law,

$$\begin{gathered} \sum f_x=Tsin\theta =ma.....\left(1\right) \\ \sum f_y=Tcos\theta -mg=0.......\left(2\right) \end{gathered}$$

$$\begin{gathered} Tsin\theta =0.5kg\times 3.00{\text{m/s}}^2 \\ =1.5kg·{\text{m/s}}^2 \\ =1.5N \end{gathered}$$

$$\begin{gathered} Tcos\theta =0.5kg\times 9.8{\text{m/s}}^2 \\ =4.9N \end{gathered}$$

$$\begin{gathered} tan\theta =\frac{1.5N}{4.9N} \\ \theta =ta{n}^{-1}\left(0.306\right) \\ \theta =17.0° \end{gathered}$$

The string forms a$17.0°$ degree angle with the vertical.

Calculation of the Tension in the string

(b)

We get this by substituting the value of theta into equation (1).

$$\begin{gathered} T=\frac{ma}{sin\theta } \\ =\frac{(0.5kg)\left(3.00{\text{m/s}}^2\right)}{sin17°} \\ =5.13N \end{gathered}$$

The string has a tension of$5.13N$