Category Archives: Sciences

Includes Engineering Sciences such as Statics, Dynamics, Strength of Materials, Physics, Fluid Mechanics, Chemistry

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}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, $y_0$ of the rock is 105 meters, and the initial velocity, $v_0$ is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned} \Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\ &=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\ &=0+11.036\ \text{m} \\ &=11.04 \ \text{m} \end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned} \text{y}&=\text{y}_{0}-\Delta \text{y} \\ &=105\ \text{m}-11.04\ \text{m} \\ &=93.96 \ \text{m} \end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned} \text{y} & =\frac{1}{2}\text{at}^{2} \\ &\text {Solving for t, we have}\\ \text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\ &=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\ &=4.63 \ \text{s} \end{aligned}

Therefore, to move out the hiker has about

\begin{aligned} \text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\ &=3.13\ \text{s} \end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\ \end{aligned}

we have

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\ \text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\ \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \end{aligned}

Substituting the known values,

\begin{aligned} \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\ \text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\ \text{v}_0&= {\color{green}7.00 \ \text{m/s}} \end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so $y$ is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s².

\begin{aligned} \text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\ 0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\ 0 & =7\text{t}-4.905\text{t}^{2}\\ 7\text{t}-4.905\text{t}^{2}&=0 \\ \text{t}\left( 7-4.905\text{t} \right) & =0 \\ \text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\ \end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned} 7-4.905\text{t} &=0 \\ 4.905\text{t} & = 7 \\ \text{t} & =\frac{7}{4.905} \\ \text{t}&={\color{green}1.43 \ \text{s}} \end{aligned}

The kangaroo is about 1.43 seconds long in the air.

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 3.2 – 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}

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?

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

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:

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:

Coplanar Force Systems

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Three-Dimensional Force Systems

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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.