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Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-3
Solution:
Part A
\begin{align*}
\text{Mg/ms} & = \frac{10^3 \ \text{kg}}{10^{-3} \ \text{s}} \\
& = 10^6 \text{kg/s}\\
& = \text{Gg/s}
\end{align*}Part B
\begin{align*}
\text{N/mm} & = \frac{1\ \text{N}}{10^{-3} \ \text{m}}\\
& = 10^3 \ \text{N/m}\\
& = \text{kN/m}
\end{align*}\begin{align*}
\frac{\text{mN}}{\left( \text{kg} \cdot \mu \text{s} \right)} & = \frac{10^{-3} \ \text{N}}{10^{-6} \ \text{kg} \cdot \text{s}}\\
& =10^3 \ \text{N}/\left( \text{kg} \cdot \text{s} \right)\\
& = \text{kN}/\left( \text{kg} \cdot \text{s}\right)
\end{align*}
We need to find the angle that force T makes with the positive x-axis first. We call this the angle beta, β. This is depicted in the free-body diagram.
Free-body diagram:
Solving for the values of angles α and β.
\begin{aligned}
\tan \alpha & = \dfrac{3}{4} \\
\alpha & = \tan ^{-1} \frac{3}{4} \\
\alpha & = 36.8699 \degree \\
\end{aligned}Knowing that the sum of angles α and β is 90°, we can solve for the β.
\begin{aligned}
\alpha + \beta & = 90\degree \\
\beta & = 90 \degree - \alpha \\
\beta & = 90 \degree - 36.8699 \degree \\
\beta & = 53.1301 \degree
\end{aligned}Equations of Equilibrium:
Summation of forces in the x-direction:
\begin{aligned}
\xrightarrow{+} \sum F_x & = 0 \\
T \cos \beta - \frac{4}{5} F & = 0 \\
T \cos 53.1301 \degree - \frac{4}{5} F & = 0 & & \qquad \qquad (1)\\
\end{aligned}Summation of forces in the y-direction:
\begin{aligned}
+\uparrow \sum F_y & =0 \\
9 - \frac{3}{5} F- T \sin \beta & = 0 \\
T \sin 53.1301 \degree + \frac{3}{5}F & = 9 & & \qquad \qquad(2)\\
\end{aligned}Now, we have two equations with two unknowns. We shall solve the unknowns by solving these equations simultaneously. We can use our calculator, or we can solve this manually using the method of substitution.
Using equation (1), solve for T in terms of F.
\begin{aligned}
T \cos 53.1301\degree-\frac{4}{5} F & = 0 \\
T \cos 53.1301\degree & = \frac{4}{5} F \\
T & = \dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \qquad \qquad (3)\\
\end{aligned}Now, substitute this equation (3) to equation (2) to solve for F:
\begin{aligned}
T \sin 53.1301 \degree + \frac{3}{5}F & = 9 \\
\left(\dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \right) \sin 53.1301 \degree + \frac{3}{5}F & = 9 \\
\frac{4}{5}F \left( \dfrac{\sin 53.1301\degree}{\cos 53.1301\degree}\right)+ \frac{3}{5}F & = 9 \\
\frac{4}{5}F \tan 53.1301\degree+\frac{3}{5}F &=9 \\
F\left( \frac{4}{5} \cdot \tan 53.1301\degree +\frac{3}{5}\right) & = 9\\
F & = \dfrac{9}{\frac{4}{5} \cdot \tan 53.1301\degree +\frac{3}{5}} \\
F & = 5.4 \ \text{kN} \\
\end{aligned}Substitute the value of F to equation (3) to solve for T:
\begin{aligned}
T & = \dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \\
T & = \dfrac{\frac{4}{5} \cdot \left( 5.4 \ \text{kN}\right)}{\cos 53.1301\degree} \\
T & = 7.2 \ \text{kN}
\end{aligned}Therefore, F = 5.4 \ \text{kN} and T= 7.2 \ \text{kN} .
Free-body diagram of the roller:
Equations of Equilibrium:
Note that if we take the sum of forces in the x-direction, there are two unknown forces involve, but if we take the sum of forces in the y-direction, there is only one unknown force involve.
Summation of forces in the y-direction:
\begin{aligned}
+\uparrow \sum F_y & =0& & & & & \\
125- N_C \cos 40 \degree &=0 & & & & &\\
N_C &=\dfrac{125}{\cos 40 \degree} & & & & & \\
N_C & =163.1759 \ \text{N} \\
\end{aligned}Summation of forces in the x-direction:
\begin{aligned}
\xrightarrow{+} \sum F_x & =0 \\
N_B - 163.1759\ \sin 40 \degree &=0 \\
N_B &=163.1759 \sin 40\degree \\
N_B & = 104.8874 \ \text{N}
\end{aligned}Therefore, the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium are 163.1759 N and 104.8874 N, respectively.
Free-body diagram:
Equations of Equilibrium:
Take the sum of horizontal forces considering forces to the right positive, and equate to zero.
\begin {aligned}
\sum{F}_x &= 0 & \\
F_1 \cos{60 \degree}+F_2 \sin{70 \degree}-5\cos{30 \degree}-\dfrac{4}{5}\left(7\right) &= 0 &\\
0.5F_1+0.9397F_2&=9.9301 &(1)\\
\end {aligned}
Take the sum of vertical forces considering upward forces positive, and equate to zero.
\begin{aligned}
\sum F_y&=0 &\\
-F_1\sin60\degree+F_2\cos70\degree+5\sin30\degree-\dfrac{3}{5}\left(7\right)&=0 &\\
-0.8660F_1+0.3420F_2&=1.7 &(2)\\
\end{aligned}Now, we have two equations with two unknowns F_1 and F_2 . So, we have a system of two equations. We can solve this using algebra, or we can directly use our calculator with this capability. The answers are
F_1=1.83 \: \text{kN}\\
F_2=9.60 \: \text{kN}Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-1
Part A: To convert the given mass in kilogram to newton force, we simply need to multiply by the acceleration due to gravity of 9.81 m/s2. We need to take into account that 1\:\text{kg m/s}^2\:=1\:\text{N} .
\begin {aligned}
8\:\text{kg} & =8\:\text{kg}\times 9.81\:\text{m/s}^2 \\
&=78.48\:\text{N}
\end {aligned}Part B: Using the same principle from Part A, we have
\begin {aligned}
0.04\:\text{kg}&=0.04\:\text{kg}\times 9.81\:\text{m/s}^2\\
&=0.3924\:\text{N}
\end {aligned}Part C: So, we are given 760 Mg (megagram). We know that 1 Mg is equivalent to 1000 kg. Therefore, 760 Mg is equal to 760,000 kg. Therefore, we have
\begin {aligned}
760\:000\:\text{kg}&=760\:000\:\text{kg}\times 9.81\:\text{m/s}^2\\
&=7\:455\:600\:\text{N}
\end{aligned}You can complete your purchase even if you do not have a Paypal account. Just click on the appropriate card on the buttons below.
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Engineering Mechanics: Statics 14th Edition by RC Hibbeler Solution Manual by Engineering-Math.org
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Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3
SOLUTION:
The parallelogram law of the force system is shown.
Consider the triangle AOB.
Using cosine law to solve for the resultant force \textbf{F}_{\text{R}}
\begin{align*}
\textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\
& =393.2 \ \text{lb}\\
& =393\:\text{lb}\\
\end{align*}The value of angle θ can be solved using sine law.
\begin{align*}
\frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\
\sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\
\theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\
\theta & = 37.89^{\circ}\\
\end{align*}Solve for the unknown angle \phi .
\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ} The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.
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