Question

Question:. An iron bolt of mass $\mathbf{65}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{g}$hangs from a string $\mathbf{35}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{cm}$long. The top end of the string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary, but is displaced horizontally $\mathbf{28}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$to the right from the previously vertical line of the string. The magnet is located to the right of the bolt and on the same vertical level as the bolt in the final configuration. (a) Draw a free-body diagram of the bolt. (b) Find the tension in the string. (c) Find the magnetic force on the bolt.

Short answer

(a) The free body diagram of the bolt is shown below.

(b) The tension in string is $1.03\mathrm{N}$..

(c) The magnetic force on the bolt is $0.81\mathrm{N}$..

Step-by-step solution

A concept and Identification of given data

The horizontal component of tension in string balances the magnetic force and the vertical component of the tension in string balances the weight of the bolt.

Consider the given data as below.

The mass of iron bolt, $m=65\mathrm{g}$

The length of string, $L=35.7\mathrm{cm}$

The horizontal displacement of bolt,$d=28\mathrm{cm}$

(a) Drawing the free body diagram of bolt

The free body diagram of the bolt is given below

The angle of the string from horizontal is given as

$\cos \theta =\frac{d}{L}$

Substitute all the values in the above equation.

$\begin{array}{rcl}cos\theta & =& \frac{28\mathrm{cm}}{35.7\mathrm{cm}}\\ \theta & =& {\cos }^{-1}\left(0.7843\right)\\ & =& 38.{34}^{\circ }\end{array}$

Therefore, the free body diagram of the bolt is shown in figure 1.

 Step 3: (b) Determination of tension in string:

The tension in string by vertical equilibrium of bolt is given as

$T\sin \theta =mg$

Here, $g$is the gravitational acceleration and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the values in the above equation.

$\begin{array}{rcl}\mathrm{Tsin}38.{34}^{\circ }& =& \frac{65\mathrm{g}}{{10}^3\mathrm{g}}\times 1\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\\ \mathrm{T}& =& \frac{0.637}{0.62}\mathrm{kg}\mathrm{m}/{\mathrm{s}}^2\\ \mathrm{T}& =& 1.03\mathrm{N}\end{array}$

Hence, the tension in string is $1.03\mathrm{N}$.

 Step 4: (c) Determination of magnetic force on the bolt

The magnetic force on the bolt by using horizontal equilibrium is given as

$F=T\cos \theta$

Substitute all the values in the above equation.

$\begin{array}{rcl}F& =& 1.03\mathrm{N}\times \cos 38.{34}^{\circ }\\ & =& 1.03\times 0.784\\ & =& 0.81\mathrm{N}\end{array}$

Hence, the magnetic force on the bolt is $0.81\mathrm{N}$.