Q46P - University Physics with Modern Physics

Question

A nail in a pine board stops a 4.9-N  hammer head from an initial downward velocity of 3.2 m/s in a distance of 0.45 cm. In addition, the person using the hammer exerts a 15-N downward force on it. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward. (a) Draw a free-body diagram for the hammer head. Identify the reaction force for each action force in the diagram. (b) Calculate the downward force $\overrightarrow{\mathbf{F}}$ exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward. (c) Suppose that the nail is in hardwood and the distance the hammer head travels in coming to rest is only 0.12 cm. The downward forces on the hammer head are the same as in part (b). What then is the force $\overrightarrow{\mathbf{F}}$ exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

Step-by-Step Solution

Answer

(a)

The reaction force is the force exerted by the nail.

(b) The downward force exerted by the hammer head is 577.4 N.

(c) The force exerted by the hammer head is 21110.56 N.

Step 1: Identification of the given data

The given data is listed below as:

 

• The weight of the nail is, $\stackrel{\rightharpoonup }{\mathrm{w}}=4.9 \mathrm{N}$
• The initial downward velocity of the hammer head is, $\mathrm{u}=3.2 \mathrm{m}/\mathrm{s}$
• The distance moved by the hammer head is, $\mathrm{s}=0.45 \mathrm{cm}\times \left(\frac{{10}^{-2}\mathrm{m}}{1 \mathrm{cm}}\right)=0.45\times {10}^{-2} \mathrm{m}$
• The weight of the person is, ${\overrightarrow{F}}_{person}=15 N$
• The distance travelled by the hammer head before coming to rest is, $d=0.12 cm\times \left(\frac{{10}^{-2} m}{1 cm}\right)=0.12\times {10}^{-2} m$

Step 2: Significance of a particle which moves under constant acceleration

The kinematic equation mainly describes the motion of a particle which moves under constant acceleration. The square of the final velocity of the particle is equal to the addition of the square of the initial velocity and twice of the product of the acceleration and the distance moved by the particle.

Step 3: (a) Determination of the free body diagram and identification reaction force for each action force

The free body diagram of the hammer head has been provided below:

In the above diagram, the force exerted by the person ${\overrightarrow{F}}_{person}$ is directed downwards along with the weight of the hammer head $\overrightarrow{w}$. Moreover, the force exerted by the nail ${\overrightarrow{F}}_{nail}$ is exerted upwards.

 

The action force is the force exerted by the hammer in the above diagram. The action force subsequently produces a reaction force which is the force exerted by the nail ${\overrightarrow{F}}_{nail}$.

 

Thus, the reaction force is the force exerted by the nail.

Step 5: (b) Determination of the downward force

The equation of the acceleration of the hammerhead is expressed as:

 

$v^2=u^2+2as$ 

 

Here, v is the final speed and u is the initial speed of the hammer head. s is the distance moved by the hammer head and a is the acceleration of the hammer head.

 

As the hammer head finally came to rest, hence the final velocity of the hammer head is zero.

 

Substitute 0 for v in the above equation.

$$\begin{gathered} 0=u^2+2as \\ a=\frac{-u^{2}s}{2s} \end{gathered}$$ 

 

The equation of the downward force exerted by the nail is expressed as:

 

${\overrightarrow{F}}_{nail}=w+{\overrightarrow{F}}_{person}-\frac{w}{g}a$ 

 

Here, ${\overrightarrow{F}}_{nail}$ is the force exerted by the nail, w is the weight of the nail,  ${\overrightarrow{F}}_{person}$ is the weight of the person, g is the acceleration due to gravity and a is the acceleration of the hammer.

 

Substitute the value of the equation (i) in the above equation.

$$\begin{gathered} {\overrightarrow{F}}_{nail}=w+{\overrightarrow{F}}_{person}-\frac{w}{g}\left(\frac{-u^2}{2s}\right) \\ =w+{\overrightarrow{F}}_{person}+\frac{w}{g}\left(\frac{u^2}{2s}\right) \end{gathered}$$ 

Substitute all the values in the above equation.

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{nail}}=4.9 \mathrm{N}+15 \mathrm{N}+\frac{4.9 \mathrm{N}}{9.81 \mathrm{m}/{\mathrm{s}}^2}\left(\frac{{\left(3.2 \mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.45\times {10}^{-2} \mathrm{m}\right)}\right) \\ =19.9 \mathrm{N}+0.49 \mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\left(\frac{{\left(10.24 {\mathrm{m}}^2/\mathrm{s}\right)}^2}{\left(9\times {10}^{-3} \mathrm{m}\right)}\right) \\ =19.9 \mathrm{N}+\left(0.49 \mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\right)\times \left(1137.7\mathrm{m}/{\mathrm{s}}^2\right) \\ =19.9 \mathrm{N}+557.5 \mathrm{N} \end{gathered}$$ 

 

Hence, further as:

 

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{nail}}=19.9 \mathrm{N}+557.5 \mathrm{N} \\ =577.4 \mathrm{N} \end{gathered}$$

 

As the magnitude of the force exerted by the nail and the force exerted by the hammer head is equal, then the force exerted by the hammer head is 577.4 N.

 

Thus, the downward force exerted by the hammer head is 557.4 N.

Step 6: (c) Determination of the force exerted by the hammer head

The equation of the acceleration of the hammer head is expressed as:

 

$v^2=u^2+2ad$ 

 

Here, v is the final speed and u is the initial speed of the hammer head. d is the distance travelled by the hammer head before coming to rest and a is the acceleration of the hammer head.

 

As the hammer head finally came to rest, hence the final velocity of the hammer head is zero.

 

Substitute 0 for v in the above equation.

$$\begin{gathered} 0+u^2+2ad \\ a=\frac{-u^2}{2d} \end{gathered}$$ 

 

The equation of the downward force exerted by the nail is expressed as:

${\overrightarrow{F}}_{nail}=w+{\overrightarrow{F}}_{person}-\frac{w}{g}a$ 

Here, ${\overrightarrow{F}}_{nail}$ is the force exerted by the nail, w is the weight of the nail, ${\overrightarrow{F}}_{person}$ is the weight of the person, g is the acceleration due to gravity and is the acceleration of the hammer.

 

Substitute the value of the equation (i) in the above equation.

$$\begin{gathered} {\overrightarrow{F}}_{nail}=w+{\overrightarrow{F}}_{person}-\frac{w}{g}\left(\frac{-u^2}{2d}\right) \\ =w+{\overrightarrow{F}}_{person}+\frac{w}{g}\left(\frac{-u^2}{2d}\right) \end{gathered}$$ 

 

Substitute all the values in the above equation.

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{nail}}=4.9 \mathrm{N}+15 \mathrm{N}+\frac{4.9 \mathrm{N}}{9.81 \mathrm{m}/{\mathrm{s}}^2}\left(\frac{{\left(3.2 \mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.12\times {10}^{-2} \mathrm{m}\right)}\right) \\ =19.9 \mathrm{N}+0.49 \mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\left(\frac{\left(10.24 {\mathrm{m}}^2/{\mathrm{s}}^2\right)}{\left(2.4\times {10}^{-3} \mathrm{m}\right)}\right) \\ =19.9 \mathrm{N}+\left(0.49\mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\right)\times \left(4266.6\mathrm{m}/{\mathrm{s}}^2\right) \\ =19.9 \mathrm{N}+2090.66 \mathrm{N} \end{gathered}$$ 

 

Hence, further as:

 

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{nail}}=19.9 \mathrm{N}+2090.66\mathrm{N} \\ =2110.56 \mathrm{N} \end{gathered}$$

 

As the magnitude of the force exerted by the nail and the force exerted by the hammer head is equal, then the force exerted by the hammer head is 2110.56 N.

 

Thus, the force exerted by the hammer head is 2110.56 N .