How can I solve these two physics problems? (equilibrium and moment)

In summary, the conversation discusses a problem with finding the correct solution for calculating the weight and torque of a force in a given scenario. The participants also discuss different methods and approaches for solving the problem, including using symbols and keeping everything symbolic until the final step. They also mention a mistake in the original equation and clarify the use of the sin and cos functions in determining the vertical component of the force.
  • #1
Tapias5000
46
10
Homework Statement
1) The non-elongated length of the spring AB is 8 m. If the block is held in the equilibrium position equilibrium position shown, determine the mass of the block at D.
2) Replace the system acting on the frame by an equivalent resultant force and specify the point, measured from B, where the line of action of the resultant intersects element BC.
Relevant Equations
1) Σfx=0, Σfy=0 and F=kx
2) Σfx=0, Σfy=0, Θ=arctan(Σfy/Σfx) and ΣMb=Mrb
Imagen1.png

I tried to solve it and I got the following is it correct?
<math xmlns=http://www.w3.org/1998/Math/MathML display=block data-is-equatio=1 data-latex=\begin{array}{l}------------------\\
Hypotenuse\ 1_{AC}\\
h_{AC}=\sqrt{6^2+6^2}\\
h_{AC}=6\sqrt{2}\\
------------------\\
Hypotenuse\ 2_{AB}\ \\
h_{AB}=\sqrt{6^2+8^2}\\
h_{AB}=10\\
------------------\\
F_r=\left(l_f-l_i\right)\cdot k\ \\
F_{AB}=\left(10-8\right)\cdot40\ \\
\left[F_{AB}=80\right]\\
------------------\\
Σfx=0\\
F_{AB}\left(\frac{8}{10}\right)-F_{AC}\left(\frac{6}{6\sqrt{2}}\right)=0\\
80\left(\frac{8}{10}\right)-F_{AC}\left(\frac{1}{\sqrt{2}}\right)=0\\
F_{AC}\left(\frac{1}{\sqrt{2}}\right)=80\left(\frac{8}{10}\right)\\
F_{AC}=80\left(\frac{8\sqrt{2}}{10}\right)\\
\left[F_{AC}=90.51\right]\\
------------------\\
Σfy=0\\
F_{AB}\left(\frac{6}{10}\right)+F_{AC}\left(\frac{6}{6\sqrt{2}}\right)-D=0\\
80\left(\frac{6}{10}\right)+90.51\left(\frac{1}{\sqrt{2}}\right)-D=0\\
D=80\left(\frac{6}{10}\right)+90.51\left(\frac{1}{\sqrt{2}}\right)\\
\left[D=112=w\right]\\
m=\frac{w}{g}\\
m=\frac{112}{9.8}\\
\left[m=11.42kg\right]\\
------------------\end{array}><mtable columnalign=left columnspacing=1em rowspacing=4pt><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>H</mi><mi>y</mi><mi>p</mi><mi>o</mi><mi>t</mi><mi>e</mi><mi>n</mi><mi>u</mi><mi>s</mi><mi>e</mi><mtext></mtext><msub><mn>1</mn><mrow data-mjx-texclass=ORD><mi>A</mi><mi>C</mi></mrow></msub></mtd></mtr><mtr><mtd><msub><mi>h</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>C</mi></mrow></msub><mo>=</mo><msqrt><msup><mn>6</mn><mn>2</mn></msup><mo>+</mo><msup><mn>6</mn><mn>2</mn></msup></msqrt></mtd></mtr><mtr><mtd><msub><mi>h</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>C</mi></mrow></msub><mo>=</mo><mn>6</mn><msqrt><mn>2</mn></msqrt></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>H</mi><mi>y</mi><mi>p</mi><mi>o</mi><mi>t</mi><mi>e</mi><mi>n</mi><mi>u</mi><mi>s</mi><mi>e</mi><mtext></mtext><msub><mn>2</mn><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mtext></mtext></mtd></mtr><mtr><mtd><msub><mi>h</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mo>=</mo><msqrt><msup><mn>6</mn><mn>2</mn></msup><mo>+</mo><msup><mn>8</mn><mn>2</mn></msup></msqrt></mtd></mtr><mtr><mtd><msub><mi>h</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mo>=</mo><mn>10</mn></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><msub><mi>F</mi><mi>r</mi></msub><mo>=</mo><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><msub><mi>l</mi><mi>f</mi></msub><mo>−</mo><msub><mi>l</mi><mi>i</mi></msub><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>⋅</mo><mi>k</mi><mtext></mtext></mtd></mtr><mtr><mtd><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mo>=</mo><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>10</mn><mo>−</mo><mn>8</mn><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>⋅</mo><mn>40</mn><mtext></mtext></mtd></mtr><mtr><mtd><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mo>=</mo><mn>80</mn><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>Σ</mi><mi>f</mi><mi>x</mi><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>B</mi></mrow></msub><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>8</mn><mn>10</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>−</mo><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>C</mi></mrow></msub><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>6</mn><mrow><mn>6</mn><msqrt><mn>2</mn></msqrt></mrow></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mn>80</mn><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>8</mn><mn>10</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>−</mo><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>A</mi><mi>C</mi></mrow></msub><

and 2
1632458891856.png

My solution... is a negative distance?
<math xmlns=http://www.w3.org/1998/Math/MathML display=block data-is-equatio=1 data-latex=\begin{array}{l}----------------------\\
hypotenuse\\
h=\sqrt{6^2+8^2}\\
h=10\\
----------------------\\
F_{rx}=Σfx\\
300\left(\frac{8}{10}\right)+100\cos\left(60\right)=Σfx\\
\left[\textcolor{#E94D40}{290=Σfx}\right]\\
F_{ry}=Σfy\\
300\left(\frac{6}{10}\right)+100\sin\left(60\right)=Σfy\\
\left[\textcolor{#E94D40}{267=Σfy}\right]\\
----------------------\\
Σr=\sqrt{290^2+267^2}\\
Σr=394.2\\
Θ=\arctan\left(\frac{267}{290}\right)\\
Θ=42.64°\\
----------------------\\
\end{array}><mtable columnalign=left columnspacing=1em rowspacing=4pt><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>h</mi><mi>y</mi><mi>p</mi><mi>o</mi><mi>t</mi><mi>e</mi><mi>n</mi><mi>u</mi><mi>s</mi><mi>e</mi></mtd></mtr><mtr><mtd><mi>h</mi><mo>=</mo><msqrt><msup><mn>6</mn><mn>2</mn></msup><mo>+</mo><msup><mn>8</mn><mn>2</mn></msup></msqrt></mtd></mtr><mtr><mtd><mi>h</mi><mo>=</mo><mn>10</mn></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>r</mi><mi>x</mi></mrow></msub><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>x</mi></mtd></mtr><mtr><mtd><mn>300</mn><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>8</mn><mn>10</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>+</mo><mn>100</mn><mi>cos</mi><mo data-mjx-texclass=NONE>⁡</mo><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>60</mn><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>x</mi></mtd></mtr><mtr><mtd><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mstyle mathcolor=#E94D40><mn>290</mn><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>x</mi></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><msub><mi>F</mi><mrow data-mjx-texclass=ORD><mi>r</mi><mi>y</mi></mrow></msub><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>y</mi></mtd></mtr><mtr><mtd><mn>300</mn><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>6</mn><mn>10</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>+</mo><mn>100</mn><mi>sin</mi><mo data-mjx-texclass=NONE>⁡</mo><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>60</mn><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>y</mi></mtd></mtr><mtr><mtd><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mstyle mathcolor=#E94D40><mn>267</mn><mo>=</mo><mi>Σ</mi><mi>f</mi><mi>y</mi></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>Σ</mi><mi>r</mi><mo>=</mo><msqrt><msup><mn>290</mn><mn>2</mn></msup><mo>+</mo><msup><mn>267</mn><mn>2</mn></msup></msqrt></mtd></mtr><mtr><mtd><mi>Σ</mi><mi>r</mi><mo>=</mo><mn>394.2</mn></mtd></mtr><mtr><mtd><mi>Θ</mi><mo>=</mo><mi>arctan</mi><mo data-mjx-texclass=NONE>⁡</mo><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>267</mn><mn>290</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow></mtd></mtr><mtr><mtd><mi>Θ</mi><mo>=</mo><mn>42.64</mn><mrow data-mjx-texclass=ORD><mo data-mjx-pseudoscript=true>°</mo></mrow></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr></mtable></math>
gVaQeWcfcZc2lM4XJSF1GN0NgYigqsRTo4zKkbzdZl2sU3L=s0.png


<math xmlns=http://www.w3.org/1998/Math/MathML display=block data-is-equatio=1 data-latex=\begin{array}{l}----------------------\\
ΣM_b=1000-300\left(\frac{8}{10}\right)\cdot3-100sen\left(60\right)\cdot2\\
\left[\textcolor{#E94D40}{ΣM_b=106.79lb_p}\right]\\
\left[\textcolor{#E94D40}{Mr_b=-267\cdot d}\right]\\
----------------------\\
ΣM_b=Mr_b\\
106.79lb_p=-267\cdot d\\
-0.4p=d\\
----------------------\end{array}><mtable columnalign=left columnspacing=1em rowspacing=4pt><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>Σ</mi><msub><mi>M</mi><mi>b</mi></msub><mo>=</mo><mn>1000</mn><mo>−</mo><mn>300</mn><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mfrac><mn>8</mn><mn>10</mn></mfrac><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>⋅</mo><mn>3</mn><mo>−</mo><mn>100</mn><mi>s</mi><mi>e</mi><mi>n</mi><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>60</mn><mo data-mjx-texclass=CLOSE>)</mo></mrow><mo>⋅</mo><mn>2</mn></mtd></mtr><mtr><mtd><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mstyle mathcolor=#E94D40><mi>Σ</mi><msub><mi>M</mi><mi>b</mi></msub><mo>=</mo><mn>106.79</mn><mi>l</mi><msub><mi>b</mi><mi>p</mi></msub></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mstyle mathcolor=#E94D40><mi>M</mi><msub><mi>r</mi><mi>b</mi></msub><mo>=</mo><mo>−</mo><mn>267</mn><mo>⋅</mo><mi>d</mi></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr><mtr><mtd><mi>Σ</mi><msub><mi>M</mi><mi>b</mi></msub><mo>=</mo><mi>M</mi><msub><mi>r</mi><mi>b</mi></msub></mtd></mtr><mtr><mtd><mn>106.79</mn><mi>l</mi><msub><mi>b</mi><mi>p</mi></msub><mo>=</mo><mo>−</mo><mn>267</mn><mo>⋅</mo><mi>d</mi></mtd></mtr><mtr><mtd><mo>−</mo><mn>0.4</mn><mi>p</mi><mo>=</mo><mi>d</mi></mtd></mtr><mtr><mtd><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo><mo>−</mo></mtd></mtr></mtable></math>
f4UF7r88IKBjA-7gYSwOVJXwWKfJ7tEPGF5A6Y3lmQFSqBr=s0.png
 
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  • #2
I confirm both your answers, but there are easier ways.
1. The horizontal and vertical components of the force in AC must be equal, so you can write the weight equals ##F_{AB}(\frac 35+\frac 45)##.
2. Suppose the point is x to the left of B and write that the net torque about that point is zero:
##-\frac 453(300)+\frac 35x(300)+1000+\frac{\sqrt 3}22(100)(x-2)=0##.

Whoops, typo. Should be ##-\frac 453(300)+\frac 35x(300)+1000+\frac{\sqrt 3}2(100)(x-2)=0##

As a matter of general style, I strongly recommend keeping everything symbolic until the final step, creating symbols as necessary for given numerical values. It has many advantages, including, often, less arithmetic and greater precision.
Ok, I violated my own principle just above, but if you had created symbols for the forces and lengths I would have preferred to use them.
 
Last edited:
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  • #3
haruspex said:
I confirm both your answers, but there are easier ways.
1. The horizontal and vertical components of the force in AC must be equal, so you can write the weight equals ##F_{AB}(\frac 35+\frac 45)##.
2. Suppose the point is x to the left of B and write that the net torque about that point is zero:
##-\frac 453(300)+\frac 35x(300)+1000+\frac{\sqrt 3}22(100)(x-2)=0##.

As a matter of general style, I strongly recommend keeping everything symbolic until the final step, creating symbols as necessary for given numerical values. It has many advantages, including, often, less arithmetic and greater precision.
Ok, I violated my own principle just above, but if you had created symbols for the forces and lengths I would have preferred to use them.
bruh, I can barely understand what I'm doing enough to apply other formulas... sad.

in 3/5*x*300 generates angular momentum at b?
also by doing your formula I get a different value at x and it is also positive.
 
  • #4
Tapias5000 said:
in 3/5*x*300 generates angular momentum at b?
No, but we are interested in the moment about a point distance x to the left of B. Both horizontal and vertical components of the 300lb force can have a moment about that.
Tapias5000 said:
also by doing your formula I get a different value at x and it is also positive.
I made a typo in the LaTeX, doubling up a 2. See the corrected post.
 
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  • #5
haruspex said:
No, but we are interested in the moment about a point distance x to the left of B. Both horizontal and vertical components of the 300lb force can have a moment about that.

I made a typo in the LaTeX, doubling up a 2. See the corrected post.
oooh, indeed the result is the same, is a negative distance logical?
how do you know that you can equal it to 0? do you have any video where you explain this method?
how do you get that sqrt(3)/2, and how do you know where to put x?

Sorry, this is the first time I see this method...
 
  • #6
Tapias5000 said:
oooh, indeed the result is the same, is a negative distance logical?
how do you know that you can equal it to 0? do you have any video where you explain this method?
how do you get that sqrt(3)/2, and how do you know where to put x?

Sorry, this is the first time I see this method...
I understand that sin(60)=sqrt(3)/2
but cos(60)=1/2 how do you get a factor of sqrt(3)/2 then?
 
  • #7
Tapias5000 said:
oooh, indeed the result is the same, is a negative distance logical?
how do you know that you can equal it to 0? do you have any video where you explain this method?
how do you get that sqrt(3)/2, and how do you know where to put x?

Sorry, this is the first time I see this method...
Suppose the equivalent force is F passing through the line BC at point P. The force has no moment about P, so the system of forces has no moment about P.

It certainly could turn out to be to the right of B. We know F's vertical component must be up the page, so the stronger the anticlockwise applied torque at C, the further to the right F must act.
 
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  • #8
Tapias5000 said:
I understand that sin(60)=sqrt(3)/2
but cos(60)=1/2 how do you get a factor of sqrt(3)/2 then?

It's the vertical component of the force at C. Since the horizontal component acts along BC, it can have no moment about P, but the vertical component will be distance x-2 to the left of P.
 
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  • #9
ooh okay, i think I get the idea
 
  • #10
haruspex said:
It's the vertical component of the force at C. Since the horizontal component acts along BC, it can have no moment about P, but the vertical component will be distance x-2 to the left of P.
thank you for the clarifications.
 
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FAQ: How can I solve these two physics problems? (equilibrium and moment)

1. How do I determine the forces in equilibrium for a given system?

To determine the forces in equilibrium for a given system, you must first draw a free body diagram of the system. Then, apply Newton's Second Law (F=ma) to each individual object in the system. The sum of all forces acting on each object should equal zero for the system to be in equilibrium.

2. How do I calculate the moment of a force?

The moment of a force is calculated by multiplying the magnitude of the force by the perpendicular distance from the point of rotation to the line of action of the force. This can also be written as M = F x d, where M is the moment, F is the force, and d is the distance.

3. What is the difference between static and dynamic equilibrium?

Static equilibrium occurs when all forces acting on a system are balanced and the system is at rest. Dynamic equilibrium, on the other hand, occurs when the system is in motion at a constant velocity, with all forces and torques balanced.

4. How can I determine the center of gravity for a given object?

The center of gravity for an object can be determined by finding the point at which the entire weight of the object can be considered to act. This can be done by balancing the object on a pivot point and marking the point directly below the pivot as the center of gravity.

5. What are some common mistakes to avoid when solving equilibrium and moment problems?

Some common mistakes to avoid when solving equilibrium and moment problems include forgetting to include all forces and torques in the free body diagram, using incorrect units, and not considering the direction of forces and torques. It is also important to double check all calculations and to make sure the system is truly in equilibrium before moving on to solving for unknown variables.

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