Author Archives: Engineering Math

Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is μs=0.60\mu _s=0.60; (b) the pavement is icy and μs=0.25\mu _s=0.25.

Solution:

The forces on the car are gravity mg downward, the normal force FN exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

∑Fy=m acFN − mg=0FN=mgFN=(1000 kg)(9.81 m/s2)FN=9810 N\begin{align*} \sum_{}^{}F_y & =m\ a_c\\ \\ F_N\ -\ mg & =0\\ \\ F_N&=mg\\ \\ F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\ \\ F_N &=9810\ \text{N} \end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

∑Fc=m ac=m⋅v2r=(1000 kg)⋅(15 m/s)250 m=4500 N\begin{align*} \sum_{}^{}F_c & =m\ a_c\\ \\ & =m\cdot \frac{v^2}{r}\\ \\ & =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\ \\ &=4500\ \text{N} \end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), μs=0.60\mu _s=0.60, and the maximum friction force attainable is

∑Ffrmax=μs FN=(0.60)(9810 N)=5886 N\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\ \\ &=5886\ \text{N} \end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

∑Ffrmax=μs FN=(0.25)(9810 N)=2452.5 N\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\ \\ &=2452.5\ \text{N} \end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.

Revolving Ball (Vertical Circle) – Uniform Circular Motion Example Problem

A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).

Solution:

The ball moves in a vertical circle and is not undergoing uniform circular motion. The radius is assumed constant, but the speed v changes because of gravity. Nonetheless, the equation for centripetal acceleration ac=v2r\text{a}_\text{c} = \frac{\text{v}^2}{\text{r}} is valid at each point along the circle, and we use it at the top and bottom points. The free body diagram is shown in the figure below for both positions.

Part A

At the top (point 1), two forces act on the ball: mg, the force of gravity (or weight), and FT1, the tension force the cord exerts at point 1. Both act downward, and their vector sum acts to give the ball its centripetal acceleration ac. We apply Newton’s second law, for the vertical direction, choosing downward as positive since the acceleration is downward (toward the center):

∑Fv=macFT1 + mg=m⋅v12r\begin{align*} \sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\ \\ \text{F}_\text{T1}\ +\ \text{mg}&= \text{m} \cdot \frac{\text{v}_1^2}{\text{r}} \end{align*}

From this equation we can see that the tension force FT1at point 1 will get larger if v1 (ball’s speed at top of circle) is made larger, as expected. But we are asked for the minimum speed to keep the ball moving in a circle. The cord will remain taut as long as there is tension in it. But if the tension disappears (because v1 is too small) the cord can go limp, and the ball will fall out of its circular path. Thus, the minimum speed will occur if FT1 = 0 (the ball at the topmost point), for which the equation above becomes

mg=m⋅(v1)2r\text{mg}=\text{m}\cdot \frac{\left( \text{v}_1 \right)^2}{\text{r}}

We solve for v1, we have

v1=gr=(9.81 m/s2)(1.10 m)=3.285 m/s\begin{align*} \text{v}_1&=\sqrt{\text{gr}} \\ \\ &=\sqrt{\left( 9.81\ \text{m/s}^2 \right)\left( 1.10\ \text{m} \right)} \\ \\ &=3.285 \ \text{m/s} \end{align*}

Therefore, the minimum speed at the top of the circle if the ball is to continue moving in a circular path is about 3.285 m/s.

Part B

When the ball is at the bottom of the circle, the cord exerts its tension force FT2 upward, whereas the force of gravity, mg still acts downward. Choosing upward as positive, Newton’s second law gives:

∑Fv=macFT2 − mg=m⋅v22r\begin{align*} \sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\ \\ \text{F}_\text{T2}\ -\ \text{mg}&= \text{m} \cdot \frac{\text{v}_2 ^2}{\text{r}} \end{align*}

The speed v2 is given as twice that in (a). We solve for FT2

FT2=m⋅v2r+mg=(0.150 kg)⋅(2×3.285 m/s)21.10 m+(0.150 kg)(9.81 m/s2)=7.358 N\begin{align*} F_\text{T2} & = m\cdot \frac{v^2}{r}+mg\\ \\ & = \left( 0.150\ \text{kg} \right)\cdot \frac{\left( 2\times 3.285 \ \text{m/s}\right)^2}{1.10\ \text{m}}+\left( 0.150\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\ \\ &=7.358 \ \text{N} \end{align*}

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 FT 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 ϕ=0 \phi = 0 from the figure. Then FT 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,

v=2⋅2πrt=4πrt=4π(0.600 m)1 s=7.54 m/s\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

Fc=mac=m⋅v2r=(0.150 kg)⋅(7.54 m/s)20.600 m=14.2 N\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,

v=2⋅2πrt=4πrt=4π(0.600 m)1 s=7.54 m/s\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, ac.

ac=v2r=(7.54 m/s)20.600 m=94.8 m/s2\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:

rate=lengthtime=90.4 cm1 day=90.4 cmday×10,000 μm1 cm×1 day86400 s=10.5 μm/s\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
Continue reading →

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.

tan⁡α=34α=tan⁡−134α=36.8699°\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.

α+β=90°β=90°−αβ=90°−36.8699°β=53.1301°\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:

→+∑Fx=0Tcos⁡β−45F=0Tcos⁡53.1301°−45F=0(1)\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:

+↑∑Fy=09−35F−Tsin⁡β=0Tsin⁡53.1301°+35F=9(2)\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.

Tcos⁡53.1301°−45F=0Tcos⁡53.1301°=45FT=45Fcos⁡53.1301°(3)\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:

Tsin⁡53.1301°+35F=9(45Fcos⁡53.1301°)sin⁡53.1301°+35F=945F(sin⁡53.1301°cos⁡53.1301°)+35F=945Ftan⁡53.1301°+35F=9F(45⋅tan⁡53.1301°+35)=9F=945⋅tan⁡53.1301°+35F=5.4 kN\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:

T=45Fcos⁡53.1301°T=45⋅(5.4 kN)cos⁡53.1301°T=7.2 kN\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 kN F = 5.4 \ \text{kN} and T=7.2 kN 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:

+↑∑Fy=0125−NCcos⁡40°=0NC=125cos⁡40°NC=163.1759 N\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:

→+∑Fx=0NB−163.1759 sin⁡40°=0NB=163.1759sin⁡40°NB=104.8874 N\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:

→+ ∑Fx=05 kN+Fsin⁡θ−8 kNcos⁡30°−4 kNcos⁡60°=0Fsin⁡θ=3.9282(1)\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:

+↑∑Fy=08sin⁡30°−4sin⁡60°−Fcos⁡θ=0Fcos⁡θ=0.5359(2)\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)

Fsin⁡θFcos⁡θ=3.92820.5359sin⁡θcos⁡θ=7.3301\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⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta} :

tan⁡θ=7.3301θ=tan⁡−17.3301θ=82.2°\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

Fsin⁡82.2°=3.9282F=3.96 kN\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:

∑Fx=06sin⁡70°+F1cos⁡θ−5cos⁡30°−45(7)=0F1cos⁡θ=4.2920(1)\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:

∑Fy=06cos⁡70°+5sin⁡30°−F1sin⁡θ−35(7)=0F1sin⁡θ=0.3521(2)\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 F1 F_1 and θ \theta . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for F1 F_1 in terms of θ \theta .

F1cos⁡θ=4.2920F1=4.2920cos⁡θ(3)\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).

F1sin⁡θ=0.3521(4.2920cos⁡θ)sin⁡θ=0.35214.2920⋅sin⁡θcos⁡θ=0.35214.2920tan⁡θ=0.3521tan⁡θ=0.35214.2920θ=tan⁡−10.35214.2920θ=4.69°\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 θ \theta to equation (3).

F1=4.2920cos⁡θF1=4.2920cos⁡4.69°F1=4.31kN\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:

F1= 4.31 kNθ= 4.69°\begin{aligned} F_1= & \:4.31 \: \text {kN} \\ \theta = & \: 4.69 \degree \end{aligned}