Seemingly simple equilibrium problem - force balance

[davidwinth]

Homework Statement

Find the cable tension and the compressive force in the beam

Relevant Equations

Equilibrium means sum of forces equals zero

For this problem we are asked to find the tension in the cable BE and the compressive force in beam CE. We are given that $\angle C = 40 \circ$. We are also given that CE = 10 meters and vertical BC = 6 meters.

My solution is to find BE using the law of cosines, from which I get

$$BE = \sqrt{10^2 + 6^2 - 2(6)(10)cos(40)} = 6.638875 m$$

I use the law of sines to calculate the angle E (taken as CEB) as

$$ \angle E = sin^{-1} \left ( \frac{6sin(40)}{6.638875} \right) = 35.51602 \circ$$

Now the sum of the forces in the x direction at point E is just (note the angle from the mass to E to B is 75.51602 degrees):

$$-F_{BE}sin(75.51602) + F_{CE}sin(40)=0$$

and the sum of the forces in the y direction is:

$$-F_{BE}cos(75.51602) + F_{CE}cos(40)=200*9.8$$

When I solve this I get that the cable tension is 2168.7 N and the compressive force in the beam is 3266.7 N. The textbook has a value for the cable of 2450 N and the beam of 3360 N. I am not sure why my solution is wrong.

 

[Steve4Physics]

Homework Statement:: Find the cable tension and the compressive force in the beam
Relevant Equations:: Equilibrium means sum of forces equals zero

For this problem we are asked to find the tension in the cable BE and the compressive force in beam CE. We are given that $\angle C = 40 \circ$. We are also given that CE = 10 meters and vertical BC = 6 meters.

My solution is to find BE using the law of cosines, from which I get

$$BE = \sqrt{10^2 + 6^2 - 2(6)(10)cos(40)} = 6.638875 m$$

I use the law of sines to calculate the angle E (taken as CEB) as

$$ \angle E = sin^{-1} \left ( \frac{6sin(40)}{6.638875} \right) = 35.51602 \circ$$

Now the sum of the forces in the x direction at point E is just (note the angle from the mass to E to B is 75.51602 degrees):

$$-F_{BE}sin(75.51602) + F_{CE}sin(40)=0$$

and the sum of the forces in the y direction is:

$$-F_{BE}cos(75.51602) + F_{CE}cos(40)=200*9.8$$

When I solve this I get that the cable tension is 2168.7 N and the compressive force in the beam is 3266.7 N. The textbook has a value for the cable of 2450 N and the beam of 3360 N. I am not sure why my solution is wrong.View attachment 291027

I did a quick check and get the same as you, I guess the official answer is wrong - it sometimes happens.

Also, note if you are using g=9.8m/s², the answers should be rounded to two sig. figs. =- you can get away with three, but five is too many!

 

[davidwinth]

I did a quick check and get the same as you, I guess the official answer is wrong - it sometimes happens.

Also, note if you are using g=9.8m/s², the answers should be rounded to two sig. figs. =- you can get away with three, but five is too many!

Thanks. If you did the problem and got my answer then I guess the book is wrong. I generally include the amount of figures that looks good to me. The "significant figures" rules never made much sense to me.

 

[haruspex]

It gets a lot closer to the given answer if we take 40° as the angle to the horizontal, but still a bit off - about 2% too high. Making g 10 instead of 9.8 doesn't help as that adds another 2%, not subtracts it.

I strongly recommend working entirely symbolically until the end. Create variable names for all the given numbers. It has more readable, checkable, correctible and often less work with a more accurate answer.

The sig figs rules only work, generally, if all the operations are multiplicative (includes divisions). In this problem there is an addition/subtraction, which can reduce the valid number of valid digits in the answer, potentially losing the lot!, but won't add any.

 

[Steve4Physics]

The "significant figures" rules never made much sense to me.

Failure to correctly apply significant figures rules may lead to a joke...

A man is admiring a dinosaur skeleton in a museum. A curator comes up and says “That one is two hundred million and one years old.”.

“Really?” says the man. “How can you tell the age so precisely?”.

“Well,” replies the curator, “when it was delivered one year ago, I was told it was two hundred million years old,”.

 

[davidwinth]

Failure to correctly apply significant figures rules may lead to a joke...

A man is admiring a dinosaur skeleton in a museum. A curator comes up and says “That one is two hundred million and one years old.”.

“Really?” says the man. “How can you tell the age so precisely?”.

“Well,” replies the curator, “when it was delivered one year ago, I was told it was two hundred million years old,”.

Funny!

However, it may be more a matter of unneeded (ridiculous) precision than of significant figures. For example, my daughter was born on May 3, 2019 at 3:23 pm. When asked how old she is, I say 2 and a half, even though I could give her age down to the years, months, days, and hours (maybe minutes) with justified accuracy. Nobody cares about the days, hours or minutes of the age of a tiddler. Similarly, 1 year out of 200 million, even if the 200 million were exact, doesn't change much worth the time to speak it.If the calculations given in the question above were for comparison with the rated strength of the cable, say, the I'd round up anyway. I'd report that the cable is under 2200 N load and use that to see how close it is to breaking.

 

[Steve4Physics]

However, it may be more a matter of unneeded (ridiculous) precision than of significant figures. For example, my daughter was born on May 3, 2019 at 3:23 pm. When asked how old she is, I say 2 and a half, even though I could give her age down to the years, months, days, and hours (maybe minutes) with justified accuracy. Nobody cares about the days, hours or minutes of the age of a tiddler. Similarly, 1 year out of 200 million, even if the 200 million were exact, doesn't change much worth the time to speak it.If the calculations given in the question above were for comparison with the rated strength of the cable, say, the I'd round up anyway. I'd report that the cable is under 2200 N load and use that to see how close it is to breaking.

Context is important. In science/engineering, the number of significant figures caries useful information. For example:

Stating a value (of x say) as x=1.34 (three significant figures) implies the value of x is such that
1.335 ≤ x < 1.345

But stating x=1.340 (four significant figures) implies the value of x is such that
1.3395 ≤ x < 1.3405

So, for example, if you have one more significant figure than appropriate, you are effectively claiming that you know the precision ten times better than you really do.

Of course, if needed, a value and its uncertainty can be formally specified, e.g. x = 1.34±0.07

But when only an informal indication of uncertainty is needed, correct use of significant figures is common practice.

And in many examinations, if you give a final answer to a numerical question to an inappropriate number of significant figures, you may lose a mark!