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Q81PE

Expert-verifiedFound in: Page 399

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Suppose you hit a steel nail with a \({\rm{0}}{\rm{.500 - kg}}\) hammer, initially moving at \({\rm{15}}{\rm{.0 m/s}}\) and brought to rest in \({\rm{2}}{\rm{.80 mm}}\). (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is \({\rm{2}}{\rm{.50 mm}}\) in diameter and \({\rm{6}}{\rm{.00 - cm}}\) long? (c) What pressure is created on the \({\rm{1}}{\rm{.00 - mm}}\)-diameter tip of the nail?**

(a) The average force exerted on the nail is obtained as:\(F = 20089.285 N\).

(b) The nail compressed is obtained as:\(\Delta L = 1.125 mm\).

(c) The pressure is obtained as:\(P = 2.56 \times 1{0^{10}} N/{m^2}\).

**Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.**

Mass of the hammer is 1.05 kg.

Initial velocity of hammer is u = 15 m/s.

Final velocity of hammer is v = 0 m/s.

(a) Displacement of hammer is:

\(\begin{array}{c}s = 2.80 mm\\ = 2.80 \times 1{0^{ - 3}} m\end{array}\)

Diameter of nail is:

\(\begin{array}{c}D = 2.55 m\\ = 2.55 \times 1{0^{ - 3}} m\end{array}\)

Radius of nail is:\(r = 1.275 \times 1{0^{ - 3}} m\)

Length of nail is:

\(\begin{array}{c}L = 6 cm \\ = 6 \times 1{0^{ - 2}} m\end{array}\)

Diameter of tip of nail is:

\(\begin{array}{c}D tip = 1 mm \\ = 1 \times 1{0^{ - 3}} m\end{array}\)

Radius of tip of the nail is:\({r_{tip}} = 0.5 \times 1{0^{ - 3}} m\)

Young’s modulus of steel is:\(Y = 2.10 \times 1{0^{11}}\)

Average force exerted on the nail is: F =?

Compression of the nail is: ΔL = ?

The pressure exerted on the tip of the nail is P =?

The force generated by the hammer is given by:

\(F = ma\)

The acceleration of the hammer is given by the kinematic equation.

\({v^2} = {u^2} + 2as\)

\(\begin{array}{l}a = - {u^2}/2s\\a = - 1{5^2}/(2*2.80 \times 1{0^{ - 3}})\\a = - 225/(5.6 \times 1{0^{ - 3}})\\a = - 40178.57m/{s^2}\end{array}\)

Here, the -ve sign gives the direction of acceleration.

Let’s consider magnitude only.

Thus, force becomes:

\(\begin{array}{l}F = 0.5*40178.57\\F = 20089.285 N\end{array}\)

Thereforce, the average force is:\(F = 20089.285 N\).

(b) The compression of the nail is:

\(\Delta L = \frac{{F*L}}{{Y*A}}\)

\(\begin{array}{l}\Delta L = \frac{{F*L}}{{Y*(\pi {r^2})}}\\\Delta L = \frac{{20089.285*6 \times 1{0^{ - 2}}}}{{2.10 \times 1{0^{11}}*(3.14*{{(1.275 \times 1{0^{ - 3}})}^2})}}\\\Delta L = 1205.3571/1071937.125\\\Delta L = 1.125 mm\end{array}\)

Therefore, the nail compressed is: \(\Delta L = 1.125 mm\).

(c) The pressure created on the tip of the nail is given by:

\(\begin{array}{l}P = F/A\\P = F/(\pi {r^2})\\P = 20089.285/(3.14*{(0.5 \times 1{0^{ - 3}})^2})\\P = 2.56 \times 1{0^{10}} N/{m^2}\end{array}\)

Therefore, the pressure is:\(P = 2.56 \times 1{0^{10}} N/{m^2}\).

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