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

[math]

F= mg = \frac{(65kg) (9.80 m/s^2)}{2} = 32.5g N = 318.5 N

[/math] (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:

 

[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 [math]T_{1}\sin{12}[/math] (T being tension in the cable).

 

You also know that the sum of both the vertical components is equal to [math]65gN[/math] in order to be keeping the trafic lights still.

[math]2(T_{1}\sin{12})=65gN[/math]

 

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

[5614]

[math]2( T_1 sin12 ) = 65 g N[/math]

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:

 

[math]( T_1 sin12 ) + ( T_2 sin12 ) = 65 g N[/math]

Where [math]T_2[/math] 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:

[math]( T_1 cos12 ) + ( T_2 cos12 ) = 0[/math]

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:

[math]\sin12=\frac{637.65}{T_{1}}[/math]

 

Am I correct?

[5614]

If you're formula was correct then yes because this:

[math]\sin12=\frac{637.65}{T_{1}}[/math]

Rearranges to:

[math] T_1 = \frac{637.65}{sin12}[/math]

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:

 

[math]2( T_1 sin12 ) = 65 g [/math]

 

Which rearranges to

 

[math] T_1 sin12 = \frac {65g}{2} = 318.5[/math]

 

[math] T_1 = \frac{318.5}{sin12}[/math]

 

[math] T_1 = 1531.900[/math] (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:

[math]2( T_1 sin12 ) = 65 g[/math]

 

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:

 

[math]( T_1 sin12 ) + ( T_2 sin12) = 65 g[/math]

 

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

 

Does that make it clearer?

[EvoN1020v]

Yes, thank you.

[EvoN1020v]

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

 

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).

[EvoN1020v]

[math]\sum{F_{x}} = (F_{a})_{x} - (F_{b})_{x}[/math]

 

[math]Acos12 - Bcos12 = 0[/math]

 

[math]cos12(A-B) = 0[/math]

 

[math]\frac{cos12(A-B)}{cos12}[/math]

 

[math](A-B) = 0[/math]

 

[math]A = B[/math]

 

 

Let A = B = T (Tension).

 

 

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

 

[math]Asin12 + Bsin12 - 637.65N = 0[/math]

 

[math]sin12(A+B) - 637.65N = 0[/math]

 

[math]sin12(T+T) - 637.65N = 0[/math]

 

[math]2Tsin12 - 637.65N = 0[/math]

 

[math]2Tsin12 = 637.65N[/math]

 

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

 

 

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

[5614]

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

 

I just learnt this today in school' date=' and I'll post the explanation later. (Lunch time is over, so I have to head back to class). [/quote']

 

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

[EvoN1020v]

Whoops. Pardon for all the [math]\cos[/math], as they should be [math]\sin[/math]. 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).

 

 

 

 

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

 

Am I correct?

[Klaynos]

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

[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 - [[math]\mu R[/math]] 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) [math]\mu R = 0.5 * 12cos(30) = 5.20[/math] 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.