Tag Archives: Statics Help

Hibbeler Statics 14E P2.4 – Components of a Force Along Two Non-Perpendicular Axes

The vertical force $\textbf{F}$ acts downward at A on the two-membered frame. Determine the magnitudes of the two components of $\textbf{F}$ directed along the axes of AB and AC. Set $\textbf{F}$ = 500 N.

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-4}

Solution:

Draw the components of the force using the parallelogram law. Then the triangulation rule.

Solving for F_AC using sine law.

\begin{align*} \frac{\text{F}_\text{AC}}{\sin \ 45^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\ \text{F}_\text{AC} & = \frac{500 \ \text{N} \ \sin45^\circ }{\sin\ 75^\circ }\\ \text{F}_\text{AC} & =366.0254 \ \text{N}\\ \text{F}_\text{AC} &\approx 366 \ \text{N} \end{align*}

Solve for F_AB using sine law.

\begin{align*} \frac{\text{F}_\text{AB}}{\sin \ 60^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\ \text{F}_\text{AB} & = \frac{500 \ \text{N} \ \sin60^\circ }{\sin\ 75^\circ }\\ \text{F}_\text{AB} & =448.2877 \ \text{N}\\ \text{F}_\text{AB} &\approx 448 \ \text{N} \end{align*}

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Hibbeler Statics 14E P2.2 — Solving for an Unknown Force Given the Magnitude and Direction of a Resultant and Another Force

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction $\theta$ .

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-2}

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

Engineering Mechanics by RC Hibbeler Problem 2.2 Parallelogram Law — Parallelogram Law

Engineering Mechanics by RC Hibbeler Problem 2.2 Triangulation Rule — Triangulation Rule

Considering the figure of the triangulation rule, we can solve for the magnitude of $\textbf{F}$ using the cosine law.

\begin{align*} \textbf{F} & = \sqrt{700^2+500^2-2\left( 700 \right)\left( 500 \right)\cos105^{\circ}}\\ & = 959.78 \ \text{N}\\ & = 960 \ \text{N}\\ \end{align*}

Then we use the sine law to solve for the angle $\theta$ .

\begin{align*} \frac{\sin \left(90^{\circ}-\theta \right)}{700} & = \frac{\sin 105^{\circ}}{959.78}\\ \sin \left(90^{\circ}-\theta \right) & =\frac{700 \sin 105^{\circ }}{959.78}\\ 90^{\circ}-\theta & = \sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right)\\ \theta & = 90^\circ-\sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right) \\ \theta & = 90^\circ-44.79^\circ\\ \theta & = 45.2^\circ\\ \end{align*}

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Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

If $\theta = 60 \degree$ and $\textbf{F} = 450 \ \text{N}$ , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-1}

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

Considering figure (b), we can solve for the magnitude of $\textbf{F}_R$ using the cosine law.

\begin{align*} \textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\ & = 497.01 \ \text{N}\\ & = 497 \ \text{N} \end{align*}

Then we use the sine law to solve for the interior angle $\theta$ .

\begin{align*} \frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\ \sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\ \theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\ & \text{This is an ambiguous case }\\ \theta & = 84.81^\circ \ or \ \theta =95.19^\circ \\ \end{align*}

In here, the correct angle measurement is $\theta = 95.19^{\circ}$ .

Thus, the direction angle $\phi$ of $\textbf{F}_R$ measured counterclockwise from the positive x-axis, is

\begin{align*} \phi & = \theta +60^\circ \\ & = 95.19^\circ +60^\circ \\ & = 155^\circ \end{align*}

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Hibbeler Statics 14E P1.19 — Determine the Weight of the Column with a given Density

A concrete column has a diameter of 350 mm and a length of 2 m. If the density (mass/volume) of concrete is 2.45 Mg/m³, determine the weight of the column in pounds.

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-19}

Solution:

The density of any material is given by the formula

\text{density}=\frac{\text{mass}}{\text{volume}}

From there, we can compute for the mass as

\text{mass}=\text{density} \times \text{volume}

We can solve for mass by multiplying density by volume. The density is already given, and we can compute for the volume of the concrete column by the formula of a volume of a cylinder.

\begin{align*} \text{V} & = \pi \text{r}^2 \text{h}\\ & =\pi \left( \frac{0.35\ \text{m}}{2} \right)^2 \left( 2 \ \text{m} \right)\\ & =0.1924 \ \text{m}^3 \end{align*}

Concrete Column illustration with diameter of 350 mm or 0.35 m, and a height of 2 m

Therefore, the mass of the concrete column is

\begin{align*} \text{mass} & =\text{density} \times \text{volume}\\ & = \left( 2.45 \times 10^3 \ \text{kg/m}^3 \right)\times \left( 0.1924 \ \text{m}^3 \right)\\ & =471.44 \ \text{kg}\\ \end{align*}

Now, we can solve for the weight by multiplying the mass by the acceleration due to gravity, g.

\begin{align*} \text{Weight} & = \text{mass} \times \text{acceleration due to gravity} \\ & = 471.44 \ \text{kg} \times 9.81 \ \text{m/s}^2 \\ & = 4624.78 \ \text{N} \end{align*}

Finally, we can convert the weight in Newtons to weight in pounds.

\begin{align*} 4624.78\ \text{N} & = 4624.78\ \text{N}\times \frac{1\ \text{lb}}{4.4482\ \text{N}}\\ & = 1039.70\ \text{lb}\\ & = 1.04\times 10^3 \ \text{lb}\\ & = 1.04 \ \text{kip} \end{align*}

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Hibbeler Statics 14E P1.14 — Evaluation of expression to correct SI Units

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)², (b) (52800 ms)², and (c) [548(10⁶)]^1/2 ms.

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-14}

Solution:

Part A

\begin{align*} \left( 212 \ \text{mN} \right)^2 & = \left[ 212\times 10^{-3} \ \text{N} \right]^2 \\ & = 0.0449 \ \text{N}^2 \\ & = 4.49\times 10^{-2} \ \text{N}^2\\ \end{align*}

Part B

\begin{align*} \left( 52800 \ \text{ms} \right)^2 & = \left[ 52800\times 10^{-3} \ \text{s} \right]^2 \\ & =2788 \ \text{s}^2 \\ & = 2.79 \times 10^3 \ \text{s}^2 \end{align*}

Part C

\begin{align*} \left[ 548\left( 10^6 \right) \right]^{1/2} \ \text{ms}& =23409 \ \text{ms}\\ & =23409\times 10^{-3}\ \text{s}\\ & = 23.4\times 10^3\times 10^{-3} \ \text{s}\\ & = 23.4 \ \text{s} \end{align*}

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Hibbeler Statics 14E P1.13 — Conversion of Density from Slug per Cubic Foot to Appropriate SI Unit

The density (mass volume) of aluminum is 5.26 slug/ft³. Determine its density in SI units. Use an appropriate prefix.

_{Engineering Mechanics: Statics 14^th Edition by RC Hibbeler, Problem 1-13}

Solution:

\begin{align*} 5.26 \ \text{slug/ft}^3 & =\left( \frac{5.26 \ \text{slug}}{\text{ft}^3} \right)\left( \frac{1 \ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{14.59\ \text{kg}}{1\ \text{slug}} \right)\\ & = 2710\ \text{kg/m}^3\\ & = 2.71\times 10^3 \ \text{kg/m}^3\\ & = 2.71\ \text{Mg/m}^3 \end{align*}

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Hibbeler Statics 14E P1.12 — Evaluation of Expression to Three Significant Figures with Appropriate SI Units

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (684 µm)/(43 ms), (b) (28 ms)(0.0458 Mm)/(348 mg), (c) (2.68 mm)(426 Mg).

_{Statics of Rigid Bodies 14^th Edition by RC Hibbeler, Problem 1-12}

Solution:

Part A

\begin{align*} \left( 684 \ \mu\text{m} \right)/43 \ \text{ms} & =\frac{684\times 10^{-6} \ \text{m}}{43\times 10^{-3} \ \text{s}}\\ & = \frac{15.9\times 10^{-3}\ \text{m}}{\text{s}}\\ & = 15.9 \ \text{mm/s} \end{align*}

Part B

\begin{align*} \left( 28 \ \text{ms} \right)\left( 0.0458 \ \text{Mm} \right)/\left( 348 \ \text{mg} \right) & = \frac{\left[ 28\times 10^{-3} \ \text{s} \right]\left[ 45.8\times 10^{-3}\times 10^6 \ \text{m} \right]}{348\times 10^{-3}\times 10^{-3} \ \text{kg}} \\ & = \frac{3.69\times 10^6 \ \text{m}\cdot \text{s}}{\text{kg}}\\ & = 3.69 \ \text{Mm}\cdot \text{s}/\text{kg}\\ \end{align*}

Part C

\begin{align*} \left( 2.68 \ \text{mm} \right)\left( 426 \ \text{Mg} \right) & = \left[ 2.68\times 10^{-3} \ \text{m} \right]\left[ 426\times 10^3 \ \text{kg} \right]\\ & = 1.14\times 10^3 \ \text{m}\cdot \text{kg}\\ & = 1.14 \ \text{km}\cdot \text{kg}\\ \end{align*}

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Hibbeler Statics 14E P1.10 — Representing Combinations of Units in the Correct SI Form

Represent each of the following combinations of units in the correct SI form: (a) GN∙µm, (b) kg/µm, (c) N/ks², and (d) KN/µs.

_{Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-10}

Solution:

Part A

\begin{align*} \text{GN} \cdot \mu \text{m} & = \left( 10^9 \ \text{N} \right)\left( 10^{-6} \ \text{m} \right)\\ & = 10^3 \ \text{N} \cdot \text{m}\\ & = \text{kN} \cdot \text{m} \end{align*}

Part B

\begin{align*} \text{kg/}\mu\text{m} & = \frac{10^3 \ \text{g}}{10^{-6} \ \text{m}} \\ & = 10^9 \ \frac{\text{g}}{\text{m}} \\ & = \text{Gg/m} \end{align*}

Part C

\begin{align*} \text{N/ks}^2 & = \frac{\text{N}}{\left( 10^3 \ \text{s} \right)^2}\\ & = \frac{\text{N}}{10^6 \ \text{s}^2} \\ & = 10^{-6} \ \frac{\text{N}}{\text{s}^2} \\ & = \mu \text{N}/\text{s}^2 \end{align*}

Part D

\begin{align*} \text{kN}/ \mu\text{s} & = \frac{10^3 \ \text{N}}{10^{-6} \ \text{s}} \\ & = 10^9 \ \frac{\text{N}}{\text{s}}\\ & = \text{GN/s} \end{align*}

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Hibbeler Statics 14E P1.5 — Representing a number between 0.1 and 1000 Using an Appropriate Prefix

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(10⁵) mm, (c) 0.00563 mg.

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-5}

Solution:

Part A

\begin{align*} 45 \ 320 \ \text{kN} & = 45.3 \times 10^3 \ \text{kN}\\ & = 45.3\times 10^3\times 10^3 \ \text{N}\\ & = 45.3 \times 10^6 \ \text{N}\\ & = 45.3 \ \text{MN} \end{align*}

Part B

\begin{align*} 568\left( 10^5 \right)\ \text{mm} & =568\left( 10^5 \right)\times 10^{-3} \ \text{m}\\ & = 568\times 10^2 \ \text{m} \\ & = 56.8 \times 10^3 \ \text{m}\\ & = 56.8 \ \text{km} \end{align*}

Part C

\begin{align*} 0.00563 \ \text{mg} & = 5.63\times 10^{-3} \ \text{mg}\\ & = 5.63\times 10^{-3}\times 10^{-3} \ \text{g}\\ & = 5.63\times 10^{-6} \ \text{g}\\ & = 5.63 \ \mu\text{g} \end{align*}

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Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-4}

Solution:

Part A

\begin{align*} 200\ \text{lb}\cdot \text{ft} & =\left( 200\ \text{lb}\cdot \text{ft} \right)\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\\ &=271\ \text{N}\cdot \text{m} \end{align*}

Part B

\begin{align*} 350 \ \text{lb/ft}^3& = \left( \frac{350\ \text{lb}}{1\ \text{ft}^3} \right)\left( \frac{1\ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{1\ \text{kN}}{1000\ \text{N}} \right)\\ & = 55.0\ \text{kN/m}^3 \end{align*}

Part C

\begin{align*} 8 \ \text{ft/hr}& = \left( \frac{8\ \text{ft}}{1\ \text{hr}} \right)\left( \frac{1\ \text{hr}}{3600\ \text{s}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\left( \frac{1000 \ \text{mm}}{1 \ \text{m}} \right)\\ & = 0.677\ \text{mm/s} \end{align*}

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Tag Archives: Statics Help

Hibbeler Statics 14E P2.4 – Components of a Force Along Two Non-Perpendicular Axes

The vertical force $\textbf{F}$ acts downward at A on the two-membered frame. Determine the magnitudes of the two components of $\textbf{F}$ directed along the axes of AB and AC. Set $\textbf{F}$ = 500 N.

Hibbeler Statics 14E P2.2 — Solving for an Unknown Force Given the Magnitude and Direction of a Resultant and Another Force

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction $\theta$ .

Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

If $\theta = 60 \degree$ and $\textbf{F} = 450 \ \text{N}$ , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Hibbeler Statics 14E P1.19 — Determine the Weight of the Column with a given Density

A concrete column has a diameter of 350 mm and a length of 2 m. If the density (mass/volume) of concrete is 2.45 Mg/m³, determine the weight of the column in pounds.

Hibbeler Statics 14E P1.14 — Evaluation of expression to correct SI Units

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)², (b) (52800 ms)², and (c) [548(10⁶)]^1/2 ms.

Hibbeler Statics 14E P1.13 — Conversion of Density from Slug per Cubic Foot to Appropriate SI Unit

The density (mass volume) of aluminum is 5.26 slug/ft³. Determine its density in SI units. Use an appropriate prefix.

Hibbeler Statics 14E P1.12 — Evaluation of Expression to Three Significant Figures with Appropriate SI Units

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (684 µm)/(43 ms), (b) (28 ms)(0.0458 Mm)/(348 mg), (c) (2.68 mm)(426 Mg).

Hibbeler Statics 14E P1.10 — Representing Combinations of Units in the Correct SI Form

Represent each of the following combinations of units in the correct SI form: (a) GN∙µm, (b) kg/µm, (c) N/ks², and (d) KN/µs.

Hibbeler Statics 14E P1.5 — Representing a number between 0.1 and 1000 Using an Appropriate Prefix

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(10⁵) mm, (c) 0.00563 mg.

Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

The vertical force \textbf{F} acts downward at A on the two-membered frame. Determine the magnitudes of the two components of \textbf{F} directed along the axes of AB and AC. Set \textbf{F} = 500 N.

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction \theta .

If \theta = 60 \degree and \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

A concrete column has a diameter of 350 mm and a length of 2 m. If the density (mass/volume) of concrete is 2.45 Mg/m3, determine the weight of the column in pounds.

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52800 ms)2, and (c) [548(106)]1/2 ms.

The density (mass volume) of aluminum is 5.26 slug/ft3. Determine its density in SI units. Use an appropriate prefix.

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (684 µm)/(43 ms), (b) (28 ms)(0.0458 Mm)/(348 mg), (c) (2.68 mm)(426 Mg).

Represent each of the following combinations of units in the correct SI form: (a) GN∙µm, (b) kg/µm, (c) N/ks2, and (d) KN/µs.

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, (c) 0.00563 mg.

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

The vertical force $\textbf{F}$ acts downward at A on the two-membered frame. Determine the magnitudes of the two components of $\textbf{F}$ directed along the axes of AB and AC. Set $\textbf{F}$ = 500 N.

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction $\theta$ .

If $\theta = 60 \degree$ and $\textbf{F} = 450 \ \text{N}$ , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

A concrete column has a diameter of 350 mm and a length of 2 m. If the density (mass/volume) of concrete is 2.45 Mg/m³, determine the weight of the column in pounds.

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)², (b) (52800 ms)², and (c) [548(10⁶)]^1/2 ms.

The density (mass volume) of aluminum is 5.26 slug/ft³. Determine its density in SI units. Use an appropriate prefix.

Represent each of the following combinations of units in the correct SI form: (a) GN∙µm, (b) kg/µm, (c) N/ks², and (d) KN/µs.

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(10⁵) mm, (c) 0.00563 mg.

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.