Author Archives: Engineering Math

Force on Revolving Ball (Horizontal) – Uniform Circular Motion Example Problem

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

Solution:

First we need to draw the free-body diagram for the ball. The forces acting on the ball are the force of gravity (or weight), mg downward, and the tension force F_T that the string exerts toward the hand at the center (which occurs because the person exerts that same force on the string). The free-body diagram for the ball is shown in the figure below. The ball’s weight complicates matters and makes it impossible to revolve a ball with the cord perfectly horizontal. We estimate the force assuming the weight is small, and letting $\phi = 0$ from the figure. Then F_T will act nearly horizontally and, in any case, provides the force necessary to give the ball its centripetal acceleration.

Before, we can use the formula of the centripetal force, we need to solve for the value of the linear velocity first. The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Using the formula for centripetal force, we have

\begin{align*} \text{F}_\text{c} &=\text{ma}_\text{c} \\ \\ & = \text{m} \cdot \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \left( 0.150\ \text{kg} \right) \cdot \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & = 14.2\ \text{N} \end{align*}

Therefore, the force a person must exert on a string is about 14.2 N.

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, a_c.

\begin{align*} \text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & =94.8 \ \text{m/s}^{2} \end{align*}

Physics Problem: Fast-Plant

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

Solution:

\begin{aligned} \text{rate} & = \frac{\text{length}}{\text{time}} \\ \\ &=\frac{90.4\ \text{cm}}{1 \ \text{day}}\\ \\ &=\frac{90.4\ \bcancel{\text{cm}}}{\bcancel{\text{day}}} \times\frac{10,000\ \mu m}{1\ \bcancel{\text{cm}}}\times \frac{1\ \bcancel{\text{day}}}{86400\ s}\\ \\ & = 10.5\ \mu m/s \end{aligned}

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°.

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 N_B and N_C 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 N_Band N_C 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:

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.

Solution:

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}

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 14^th 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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Author Archives: Engineering Math

Force on Revolving Ball (Horizontal) – Uniform Circular Motion Example Problem

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Physics Problem: Fast-Plant

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

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.

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°.

Solution:

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 N_B and N_C on the bearing at its contact points B and C for equilibrium.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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?

Solution:

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

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.

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°.

Solution:

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.

Solution:

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

Solution:

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.

Solution:

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

Solution:

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?

Solution:

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 N_B and N_C on the bearing at its contact points B and C for equilibrium.

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.

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