The lamp has a mass of 15 kg 12

 in Mechanics: Statics  tagged Engineering Mechanics: Statics / equilibrium / Forces

The lamp has a mass of 15 kg and is supported by a pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium.

The lamp has a mass of 15 kg

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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We will express each force in the pole and cables in Cartesian vector form. To do so, we will find the locations of the points in Cartesian vector form.

The lamp has a mass of 15 kg

Using the diagram, the locations of the points A, B, C, and O are:

A:(2i-1.5j+6k)

B:(-4i+1.5j+0k)

C:(0i+1.5j+0k)

O:(0i+0j+0k)

 

The position vectors for points from A to B, A to C, and A to O are:

r_{AB}\,=\,\left\{(-4-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-6i+3j-6k\right\}

r_{AC}\,=\,\left\{(0-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-2i+3j-6k\right\}

r_{OA}\,=\,\left\{(2-0)i+(-1.5-0)j+(6-0)k\right\}\,=\,\left\{2i-1.5j+6k\right\}

(Why did we write the position vector from O to A instead of A to O? Unlike ropes, which can only be in tension, the pole will actually be in compression. That means a force from the pole is heading upwards, where as in the cables, the force is heading away from the lamp)

A position vector, denoted \mathbf{r} is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was (x_A,y_A,z_A) and the coordinates of point B was(x_B,y_B,z_B), then r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

 

The magnitude of each position vector is:

magnitude of r_{AB}\,=\,\sqrt{(-6)^2+(3)^2+(-6)^2}\,=\,9

magnitude of r_{AC}\,=\,\sqrt{(-2)^2+(3)^2+(-6)^2}\,=\,7

magnitude of r_{OA}\,=\,\sqrt{(2)^2+(-1.5)^2+(6)^2}\,=\,6.5

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r\,=\,ai+bj+ck, then the magnitude would be, r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

 

The unit vector for each position vector is:

u_{AB}\,=\,\left(-\dfrac{6}{9}i+\dfrac{3}{9}j-\dfrac{6}{9}k\right)

u_{AC}\,=\,\left(-\dfrac{2}{7}i+\dfrac{3}{7}j-\dfrac{6}{7}k\right)

u_{OA}\,=\,\left(\dfrac{2}{6.5}i-\dfrac{1.5}{6.5}j+\dfrac{6}{6.5}k\right)

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was r\,=\,ai+bj+ck, then unit vector, u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}

 

We can now write each force in Cartesian vector form:

F_{AB}\,=\,F_{AB}\left(-\dfrac{6}{9}i+\dfrac{3}{9}j-\dfrac{6}{9}k\right)

F_{AC}\,=\,F_{AC}\left(-\dfrac{2}{7}i+\dfrac{3}{7}j-\dfrac{6}{7}k\right)

F_{OA}\,=\,F_{OA}\left(\dfrac{2}{6.5}i-\dfrac{1.5}{6.5}j+\dfrac{6}{6.5}k\right)

(Further simplify by expanding the brackets and writing the fractions in decimal form)

F_{AB}\,=\,\left\{-0.667F_{AB}i+0.333F_{AB}j-0.667F_{AB}k\right\}

F_{AC}\,=\,\left\{-0.286F_{AC}i+0.429F_{AC}j-0.857F_{AC}k\right\}

F_{OA}\,=\,\left\{0.308F_{OA}i-0.231F_{OA}j+0.923F_{OA}k\right\}

F\,=\,\left\{0i+0j-(15)(9.81)\right\}\,=\,\left\{0i+0j-147.15k\right\}

(Force F is the weight of the lamp, which only has a z-component)

 

Since the system is in equilibrium, all forces added together must equal zero.

\sum \text{F}\,=\,0

F_{AB}+F_{AC}+F_{OA}+F\,=\,0

 

Furthermore, as each force added together must equal zero, than each individual component (x, y, z-components) added together must also equal zero.

x-components:

-0.667F_{AB}-0.286F_{AC}+0.308F_{OA}\,=\,0

y-components:

0.333F_{AB}+0.429F_{AC}-0.231F_{OA}\,=\,0

z-components:

-0.667F_{AB}-0.857F_{AC}+0.923F_{OA}-147.15\,=\,0

 

Solving the three equations simultaneously gives us:

F_{AB}\,=\,110.4 N

F_{AC}\,=\,86.1 N

F_{OA}\,=\,319.2 N

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-50.

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12 thoughts on “The lamp has a mass of 15 kg”

    • Reply ↓
      questionsolutions Post author

      Hi,

      The green box answers are found by solving the component equations we found (in the blue box above). There are three equations with three unknowns and we can solve for those three. When you solve them, you will end up with the values for each force in the cables. All steps are shown to the point where you can get the three equations to solve. If you don’t remember how to solve them, please kindly look at this link: https://goo.gl/IFzQTd

      If you are still unclear, let us know, and we will try to clear it up some more!

      Thanks!

      • Reply ↓
        denny

        Eureka! Found the solutions. Thank you so very much for the assistance and guidance. This is a very lengthy problem to solve with an even longer system of equations solution.

        • Reply ↓
          questionsolutions Post author

          Glad to see you got it! Yes, it is a long and tedious task, but usually, with these types of questions, (where we don’t calculate the moment), you can follow a series of steps to arrive at the answer. They are usually as follows:
          1) Write down the locations of all the points
          2) Find the position vector for each cable/force with respect to the object
          3) Find the magnitude of each position vector
          4) Use the magnitude to find the unit vector
          5) Write each force in Cartesian vector form using the unit vector you found
          6) Solve the system of equations to find the forces

          Anyways, you are very welcome, best of luck in your studies!

    • Reply ↓
      questionsolutions Post author

      Hi. 9.81 is the acceleration due to gravity. Remember that in the question, it tells us that the mass of the lamp is 15 kg, not the weight. To find the weight, we need to multiply the mass by the acceleration due to gravity. For more information, please read: http://bit.ly/2EqvBIL

      Thank you very much 🙂

  • Reply ↓
    Garry Ingham

    A great solution. Helping me understand forces better. But I have another question in retion to this. Determine the moment about B induced by the weight of the lamp acting at A. A little help would be appreciated