Author Archives: Engineering Math

Statics 3.6 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition


The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle θ for equilibrium. The forces are concurrent at point O. Take F=8 kN.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler Problem 3-5 Equilibrium of Truss Members Connected in a Gusset Plate

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Statics 3.5 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition


The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Take θ=90°.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler Problem 3-5 Equilibrium of Truss Members Connected in a Gusset Plate

Solution:

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} .

Statics 3.4 – Normal Reactions in a Bearing | Hibbeler 14th Edition


The bearing consists of rollers, symmetrically confined within the housing. The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium.

Engineering Mechanics: Statics 14th Edition Problem 3-4 - The normal reactions on a bearing.

Solution:

Free-body diagram of the roller:

Free-body diagram of Problem 3.4 - Engineering Mechanics Statics 14th Edition by Russell C. Hibbeler| Normal Forces in a Bearing

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.


Statics 3.3 – Solving for the magnitude and direction of a force for equilibrium | Hibbeler 14th Edition


Determine the magnitude and direction θ of F so that the particle is in equilibrium.

Statics 14E Problem 3.3 Forces in Equilibrium with unknown force and its direction

Solution:

Free-body Diagram:

Equilibrium Equation:

Summation of forces in the x-direction:

\begin{aligned}
\xrightarrow{+} \: \sum F_x & = 0 & \\
5 \: \text{kN}+F \sin \theta - 8 \: \text {kN} \cos 30 \degree - 4\: \text{kN} \cos 60 \degree & = 0  & \\
F \sin \theta &= 3.9282  & (1)

\end{aligned}

Summation of forces in the y-direction:

\begin{aligned}
+\uparrow \sum F_y & = 0  &\\
8 \sin 30 \degree - 4 \sin 60 \degree - F \cos \theta & =0 &\\
F \cos \theta & = 0.5359 & (2)\\

\end{aligned}

We now have two equations. Divide Eq (1) by (2)

\begin{aligned}
\dfrac{F \sin \theta}{F \cos \theta}  &= \dfrac{3.9282}{0.5359} \\

\dfrac{ \sin \theta}{ \cos \theta} & = 7.3301 \\ 

\end{aligned}

We know that \tan \theta = \dfrac{\sin \theta}{\cos \theta} :

\begin{aligned}
\tan \theta &=7.3301 \\
\theta & = \tan^{-1}7.3301\\
\textcolor{blue}\theta & \textcolor{blue}{=82.2\degree}\\
\end{aligned}

Substituting this result to equation (1), we have

\begin{aligned}
F\sin 82.2 \degree & = 3.9282 \\
\textcolor{blue}F & \textcolor{blue}{=3.96 \ \text{kN}}
\end{aligned}

Statics 3.2 – Equilibrium of Truss Members that are Pin Connected | Hibbeler 14th Edition


The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle θ for equilibrium. Set F2=6 kN.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle
Figure 3.1/3.2

Solution:

Free-body diagram:

Free-body-diagram-for-Problem-3.2 of Engineering Mechanics: Statics by Russell C. Hibbeler

Equations of Equilibrium:

The summation of forces in the x-direction:

\begin{aligned}
\sum F_x & = 0 &\\
6 \sin 70 \degree + F_1 \cos \theta - 5 \cos 30 \degree - \dfrac {4}{5} \left(7 \right) & = 0 & \\
 F_1 \cos \theta & = 4.2920 & (1)
\end{aligned}

The summation of forces in the y-direction:

\begin{aligned}
\sum F_y & =0 & \\
6 \cos 70 \degree+5 \sin 30 \degree - F_1 \sin \theta - \dfrac{3}{5} \left( 7 \right ) & =0 & \\
F_1 \sin \theta &=0.3521 & (2)\\
\end{aligned}

We came up with 2 equations with unknowns F_1 and \theta . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for F_1 in terms of \theta .

\begin{aligned}
F_1 \cos \theta & = 4.2920  &\\
F_1 & =\dfrac{4.2920}{ \cos \theta } & (3) \\
\end{aligned}

Now, substitute this equation (3) to equation (2).

\begin{aligned}
F_1 \sin \theta & = 0.3521 \\
\left ( \dfrac {4.2920}{\cos \theta} \right) \sin \theta & =0.3521 \\
4.2920 \cdot \dfrac{\sin \theta}{\cos \theta} & = 0.3521 \\
4.2920 \tan \theta & = 0.3521 \\
\tan \theta & = \dfrac{0.3521}{4.2920} \\
\theta &= \tan ^{-1} \dfrac{0.3521}{4.2920} \\
\theta & = 4.69 \degree

\end{aligned}

Substitute the solved value of \theta to equation (3).

\begin{aligned}
F_1 & = \dfrac{4.2920}{\cos \theta} \\
F_1 &= \dfrac{4.2920}{\cos 4.69 \degree} \\
F_1 & = 4.31 \text{kN}
\end{aligned}

Therefore, the answers to the questions are:

\begin{aligned}
F_1= & \:4.31 \: \text {kN} \\
\theta = & \: 4.69 \degree
\end{aligned} 

Statics 3.1 – Equilibrium of Truss Members in Pin Connection | Hibbeler 14th Edition


The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle
Figure 3.1

Solution:

Free-body diagram:

Free-Body Diagram for Problem 3.1 of Engineering Mechanics: Statics 14th Edition by Russell C. Hibbeler

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}

Hibbeler Statics 14E P1.1 — Converting mass to weight in newtons


What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-1


Solution:

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}

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College Physics by Openstax Chapter 3 Problem 37


Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?


Solution:

Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.


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College Physics by Openstax Chapter 3 Problem 36


The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.


Solution:

We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground, and the motion started from the ground.

The formula for range is

\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}

Since we are already given the necessary details, we can now solve for the range.

\begin{align*}
 \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\
\text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ 
\end{align*}

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College Physics by Openstax Chapter 3 Problem 35


In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)


Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground.

The formula for the range is

\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.

\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump.

\begin{align*}
 \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s}
\end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

\begin{align*}
\text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\
\text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\
\text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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