Force Problem

[EvoN1020v]

I confronted this force problem in the Physics Nelson's Textbook, and my teacher gave us a lot of other questions except this one. I told her, I wanted to do it to challenge myself. But oddly enough, I'm struggling with it.

Here's the question:
A traffic light hangs in the centre of the road from cables as shown in the figure. If the mass of the traffic light is 65 kg, what force must the cables exert on the light to prevent it from falling? (Hint: Since the angles that cables make with the hortionzal are the same (12 degrees), they both exert forces of the same magntiude.

Any suggestions?

[JustStuit]

Since they must conteract the weight they would each need to pull

F= mg = \frac{(65kg) (9.80 m/s^2)}{2} = 32.5g N = 318.5 N
(at the surface of the earth) upward. Do the angles affect the force at all? I'm not sure.

:edit: I'm assuming two wires hold it up. Now I'm even less sure the angles don't mean anything.

[timo]

Draw the forces in the diagram and see if that helps you. If not, post the diagram here; someone will tell you the next step (or what you did wrong, depending on what´s the case).

[EvoN1020v]

I thought my question explained what the diagram would look like. Here's the diagram:

1212

[timo]

Yes, your question did explain it. But I thought my answer also explained that I proposed you draw the forces into that diagram - because that´s the way to solve it unless you know how to work with vectors.

EDIT: Actually, drawing the forces into the diagram is a good idea even if you know how to work with vectors. It is a great help for visualizing what you are doing.

[EvoN1020v]

I do know how to work with vectors. Fg for the hanging light? And force applied at both angles. I'm just having a problem how to find the magntiude of the force at those 2 directions.

[timo]

mg for the gravitational force acting on the light; F1 and F2 for the forces exerted by the cables. mg + F1 + F2 = m*a = 0.

[swansont]

There is no net acceleration, so all the forces must add to zero, in both the vertical and horizontal direction. Atheist has given you the equation for the vertical, and the hint in the problem tells you that F1=F2. The angle will matter if you need to find the actual tension in the cable, but that's not what was asked in the problem.

[timo]

Atheist has given you the equation for the vertical ...
Actually, I didn´t. At least, I didn´t intend to. g, F1, F2 and 0 were supposed to be vectors.

... and the hint in the problem tells you that F1=F2.
or |F1|=|F2| if they are vectors.

[the tree]

Try just looking at one cable, then draw it as the hypotenuse of a right angled triangle with the right angle being at the middle of the crossbar.

Your vertical line represents the vertical component of the cable, if you remember your trig you'll see that it's T_{1}\sin{12} (T being tension in the cable).

You also know that the sum of both the vertical components is equal to 65gN in order to be keeping the trafic lights still.
2(T_{1}\sin{12})=65gN

...and now I go look for my calculator....

[5614]

2( T_1 sin12 ) = 65 g N
Will give you the tension (as the tree said) and as both cables are at the same angle the tension in each string will be the same.

If they were at different angles then you would have to use:

( T_1 sin12 ) + ( T_2 sin12 ) = 65 g N
Where T_2 is the tension in the 2nd string. It doesn't make a difference which one you call 1 and 2, just keep it constant throughout the calculation.

You could also say that:
( T_1 cos12 ) + ( T_2 cos12 ) = 0
because as there is acceleration the forces must sum to 0, so the horizontal components of both T1 and T2 must sum to 0.

You would then combine the two equations using simultaneous equations.

You get that kind of question for Maths (Mechanics 1) if you are doing this for AS Level (pg84 (Qs 7 & 8) of the orange Heinemann Modular Mathematics Mechanics 1 book).

[EvoN1020v]

I got the value of 3,066.9 N for both of the tensions.

This is how I solve this:
\sin12=\frac{637.65}{T_{1}}

Am I correct?

[5614]

If you're formula was correct then yes because this:
\sin12=\frac{637.65}{T_{1}}
Rearranges to:
T_1 = \frac{637.65}{sin12}
Stick it in on a calculator and you get the answer 3066.927 (to 3 decimal places)

But where did you get that equation from?

We said:

2( T_1 sin12 ) = 65 g

Which rearranges to

T_1 sin12 = \frac {65g}{2} = 318.5

T_1 = \frac{318.5}{sin12}

T_1 = 1531.900 (to 3 decimal places)

Basically I think you ignored the 2 and so your answer was double the expected value. And what value did you use for g? (I've used 9.8)

[EvoN1020v]

I used the value of 9.81 for the g. Why did you divide it by 2 though? I kinda focusing my answer on one rope, not both. So you're assuring that you divide the tensions by 2 to give the value for each rope?

[5614]

I think you are asking why is there a 2 at the beginning of this equation:
2( T_1 sin12 ) = 65 g

If so it is because both ropes (ie. 2 ropes, T1 and T2) contribute a vertical tension. The vertical component of T1 AND the vertical component of T2 together sum to the total vertical tension in the whole thing. Another way of writing the formula which would be clearer is to say:

( T_1 sin12 ) + ( T_2 sin12) = 65 g

Where T1 = T2... which is why we simplified it.

Does that make it clearer?

[EvoN1020v]

Yes, thank you. :-)

[EvoN1020v]

2(T_{1}cos12)=637.65N yields 1.3x10^{3}N.

I just learnt this today in school, and I'll post the explanation later. (Lunch time is over, so I have to head back to class). :cool:

[EvoN1020v]

\sum{F_{x}} = (F_{a})_{x} - (F_{b})_{x}

Acos12 - Bcos12 = 0

cos12(A-B) = 0

\frac{cos12(A-B)}{cos12}

(A-B) = 0

A = B


Let A = B = T (Tension).


\sum{F_{y}} = (F_{a})_{y} + (F_{b})_{y} - F_{g} = 0

Asin12 + Bsin12 - 637.65N = 0

sin12(A+B) - 637.65N = 0

sin12(T+T) - 637.65N = 0

2Tsin12 - 637.65N = 0

2Tsin12 = 637.65N

T=\frac{637.65N}{2(sin12)} = 1.5x10^3N


Both of the ropes are equal, so the answer of T_{1} and T_{2} is: 1.5 x 10^{3} N.

[5614]

2(T_{1}cos12)=637.65N yields 1.3x10^{3}N.

I just learnt this today in school, and I'll post the explanation later. (Lunch time is over, so I have to head back to class). :cool:

I think you meant to put sine instead of cosine in that equation.

[EvoN1020v]

Whoops. :embarass: Pardon for all the \cos, as they should be \sin. It's too late to edit my post, so thanks 5614 for the warning.

[EvoN1020v]

I have another question:

A 1.0 kg wood block is pressed against the wood wall by the 12N force down. If the block is initially at rest, will it move upward, move downward, or stay at rest? (The coefficient of static friction for wood is 0.50 and the coefficent of kinetic friction is 0.20).

1230



My answer is that the block won't move, because the resultant vector (V_{r}) is lesser than F_{fsmax}, where V_{r} = 3.81 N, and F_{fsmax} = 5.196 N.

Am I correct?

[Klaynos]

It will stay stationary, as the gravity force is bigger than the upwards force, but this is countered by fmax

[EvoN1020v]

So you're saying that I'm correct? Are the values accurate comparing to yours?

[EvoN1020v]

Can anybody answer my question?

[5614]

We're only really worried about vertical forces here. I'll do all my working here now, hopefully it will turn out the same as yours!

All vertically:

Block pushes down with F = weight - friction = 9.8 - [\mu R] 0.5*12cos30 = 4.6...

The force supplies 12sin30 = 6N upwards. So the force supplied is enough to move the block up BUT we need to include friction for the upwards direction.

F up = 6 - 0.5*12cos30 = 0.8...

Net force vertically (take up as positive) is:
-3.6N

Remember there is friction acting against that, friction is equal to (when stationary) \mu R = 0.5 * 12cos(30) = 5.20 which is bigger than the 3.6N (direction is irrelevant here) net force we have acting on the square, therefore the square will not move.

So yeah, I get all the same values as you, your values are more accurate than mine though.

[EvoN1020v]

Thanks 5614. I got another challenging physcis question for you guys:

A 40 kg boy works at his dad's hardware store. One of the boy's jobs is to unload the delivery truck. He places each package on a 30 degrees ramp and shove it up the ramp into the storeroom. He needs to shove the package with an acceleration of at least 0.1 m/s^2 in order for the package to make it to the top of the ramp. One day the ground is wet with rain and he's wearing slick leather-soled shoes. The coefficient of static friction between his shoes and the ground is 0.25. The largest package of the day is 15 kg, and it's coefficient of static friction on the ramp is 0.40. Can he give the package a big enough shove to reach the top of the ramp without his feet slipping?

[5614]

Well lets see:

He needs to push at a=0.1

Biggest mass = 15

therefore F = ma = 15 * 0.1 = 1.5N

Now can I assume that he pushes up at 30 degrees or does his line of force act horizontally? That makes a difference. If he pushes it at 30 deg then he needs F = 1.5 but if he pushes horizontally he would need a force which gives a component of the force up the slope of 1.5N

See the difference? Does he need to supply the 1.5N or supply something which has a component of 1.5N ?

Anyway once you know the force he needs to push then remember that if he pushes the load with a force F then the load pushes him with a force F (in the opposite direction). Include friction, work out the overall force required to move the box. Use F = \mu R to see if the force he has to supply is bigger than the frictional force which is available. If friction is bigger than he can move it, if friction is smaller than the force he is applying then he will slide.

[EvoN1020v]

One thing: I mistype the value of acceleration, it should be 1.0 m/s^2 instead. Therefore, the applied force (F_{app}) would be 15 N.

First thing I did: Find the friction Force when the boy push the 15 kg box.

\sum F_x = F_{app} - F_f - (F_g)_x = ma
F_f = F_{app} - (F_g)_x - ma

F_{app} and -ma cancel each other, so that remains (F_g)_x which is mgsin30.

The friction force is: 73.575 N.

What I do next? I know that the friction force is greater than the applied force but, he have a mass of 40 kg, so I'm not sure how to join that reference to figure out my answer?

Maybe I should figure out the F_{fsmax} for his shoe to the ground? Any help here please?

[5614]

I haven't properly read your post, just the question, will reply properly tomorrow but if it helps then to involve the boy's mass... well I dunno what you did without it! (can't stop to read it now) but the boys mass will effect the maximum amount of friction he can have.

Frictional forces are F= \mu R where R is his reaction, which will be equal to his weight. Weight = mg. That's where 'm' (mass) comes into it.

Will look at the prob more tomorrow.

[EvoN1020v]

never have seen F = \mu R before?

I know that the boy's weight is 392.4 N.

[5614]

Doubley posty, oops!

[5614]

You have seen F = \mu R before! How do you work out the force which can be supplied due to friction? I think you do know it but maybe just haven't ever had it written out like that.

The force (F) is equal to the coefficient of friction (\mu) multiplied by the reaction (R).

If the boy's mass is 40kg, then his weight is 392N (I use g as 9.8) and the coefficient of friction is 0.25 then the maximum value of friction (which is a force) is F_{friction} = \mu_{coefficient-of-friction} R_{reaction} = 0.25_{given-in-the-question} * 392_{mass * gravity} = 98N

Now remember Newtons 3rd Law, if the boy pushes the box with a force F then the box pushes the boy with a force F in the opposite direction.

So now quite simply if he has to push with more than 98N then he will be being pushed with more than 98N and so he will slide, because the greatest force friction can supply is 98N. If he's pushing the box with >98N then he's being pushed with >98N which will overcome friction and so he will slide.

The problem to me is what force does he apply? Do you follow what I was saying at the end of post #27? Is he applying a force horizontally or is he applying the force up the slope? That makes a difference to the answer.

You know how to work out the frictional force on the box, it's the same method for the boy.

The maximum force due to friction on the box I get as F_{friction} = \mu R = 0.25 * 73.575_{mgsin \theta = 15 * 9.8 * sin30 = 73.5 (rounded)} = 18.394N so now you need to account for the force required to push the box (which I'm not certain on, see above) and then add the friction.

Is it getting clearer?