Physics F-U-N-doom

[EvoN1020v]

I have recieved some private messages from SFN members asking me about posting up some physics questions. Apparently my "Speed of Sound" thread went dry, so I'll try this one more time, and hopefully, people will participate.

 

The first question is a high schooler question, but it will challenge high school students, and people without formal education alike. If you can answer the question, good job. I'll put up another question as it will get harder as it goes on.

 

1st question:

A family has purchased a new 45-inch television and because they only own a subcompact car, they have placed it on the roof, as shown. Because they had no rope, the father used double-sided tape between the box and roof to help hold the box in place (this isn't an overly bright family). The TV box weighs 90 pounds, and the coefficient of static friction ([math]\mu_s[/math]) supplied by the tape is 0.9. What is the car's maximum deceleration for which the box remains in a no-slip condition? Assume a flat roof that fully supports the box.

 

 

Good luck.

[CanadaAotS]

ok... its been awhile since I've done this stuff but...

 

1 lbs = .454 kg

mass = 40.86 kg

Fg = 40.86*9.8

Fg = 400.428

 

Ff = mu*Fg

Ff = .9*400.428

Ff = 360.3852

 

Since f = ma

360.3852 = 40.86a

a = 8.82 m/s

 

Do I win a cookie?!? lol...

[EvoN1020v]

Please remain the answer in the units of feet per second squared.

[Klaynos]

Please remain the answer in the units of feet per second squared.

 

For the love of all that is wholey, use SI!!!!

 

Rather trivial really :s

[EvoN1020v]

For the love of all that is wholey, use SI!!!!

 

Rather trivial really :s

 

As an undergrad engineer, I was taught to learn the SI metric system, and the U.S. metric system. If a question have metric units, then you should have the answer with metric units. Goes same for the U.S. metric system.

 

But I did convert CanadaAots's answer, and it's right.

 

The maximum deleceration of the car is [math]29.0[/math] [math]ft/s^2[/math], or [math]8.84[/math] [math]m/s^2[/math].

 

I'll have another new problem post on Monday.

Have a good weekend.

[EvoN1020v]

For the love of all that is wholey, use SI!!!!

 

Rather trivial really :s

 

As an undergrad engineer, I was taught to learn the SI metric system, and the U.S. metric system. If a question have metric units, then you should have the answer with metric units. Goes same for the U.S. metric system.

 

But I did convert CanadaAots's answer, and it's right. I'll post my process of calculation tomorrow or Sunday.

 

The maximum deleceration of the car is [math]29.0[/math] [math]ft/s^2[/math], or [math]8.84[/math] [math]m/s^2[/math].

 

I'll have another new problem post on Monday.

Have a good weekend.

 

P.S. I don't know how to delete post #5, sorry.

[EvoN1020v]

Calculation for the 1st question:

 

Goal: Determine the maximum decleration of the car so the TV won't fall off the roof.

 

Given: [math]\mu_s=0.9[/math], [math]m_{tv} = 90lb[/math].

 

Diagram:

 

Formulate: [math]F_{net}=ma[/math]

 

Solution:

From the Free Body Diagram, you can gather all the forces into Fnet. Using the provided y and x axes, you can determine if it's in the positive and negative direction. (i_hat for x, and j_hat for y)

 

[math]-F_fi +N_{tv}j-W_{tv}j = -mai[/math]

 

Now separate the i and j:

(1) [math]i: [/math][math]-F_f = -ma[/math]

(2) [math]j: [/math][math]N_{tv}-W_{tv}=0[/math]

 

(2) equation can be rearranged to: [math]N_{tv}=W_{tv}[/math]

 

Substitute (2) in (1):

[math](-\mu(W_{tv}) = -ma[/math]

[math](-\mu(mg))=-ma[/math]

 

Insert all the numbers that you have:

[math]-0.9(90lb(32.2ft/s^2) = -(90lb)a[/math]

 

The resultant acceleration or deceleration in this situtation is [math]29.0 ft/s^2[/math].

 

 

I'll have a new problem for Monday.

[insane_alien]

Evo, this is an international forum 96% of the world uses SI. SI is the official standard of the forum so pst the questions in SI or don't complain if thats what you get the answer in.

[EvoN1020v]

Evo, this is an international forum 96% of the world uses SI. SI is the official standard of the forum so pst the questions in SI or don't complain if thats what you get the answer in.

 

I disagree with you on the 96% part. Engineers in Europe use the U.S. metric system as well.

 

I understand that you guys want it to be in the international metric system - I got no problem with that.

[Klaynos]

I disagree with you on the 96% part. Engineers in Europe use the U.S. metric system as well.

 

I understand that you guys want it to be in the international metric system - I got no problem with that.

 

Nearly all people in europe use the metric system (SI). The biggest exception to this is in the UK where most people use a combination of the two but in nearly 100% of formal exams SI is used, and anything higher than about 13yo exams for the sciences and engineering will use SI.

[insane_alien]

Evo, european engineers only deviate from SI when dealing with older machinery that was designed on imperial measures or when sending data to US engineers.

[EvoN1020v]

This question is related with conversion of momentum. Please leave the answer in i and j direction. (i.e. [math]V_b=2.3m/si -5.6m/sj[/math]). If you got the right answer, I will be really impressed, because this is a challenging university physics question.

 

2nd question:

 

A 50,000 kg submarine moving forward at 5 m/s launches a 100 kg torpedo. The ejection mechanism acts on the torpedo for 0.25s, at the end of which the torpedo enters the water with a velocity of 10 m/s relative to the submarine and at 20 degrees elevation to the orginial path of the submarine.

 

(a) Calculate the velocities of the submarine and torpedo just after ejection is complete.

 

 

Good luck.

[Klaynos]

b) calculate the impulse.

[hotcommodity]

Ok, I refrained from looking at other posts for the first question, here's what I got:

 

 

There are 3 forces acting on the object: The normal force [math]F_N[/math], the weight vector [math]W[/math], and the frictional static [math]f_s[/math]. Since acceleration only occurs in the x direction, we only consider [math]\sum F_x=ma[/math]. To find [math]f_s[/math], which is the only force acting in the x direction, I convert the mass to kilograms, and multipliy the normal force ( [math]F_N=40.9*9.8=400.82[/math] ) by the frictional coefficient of .9 to get 360.738 Newtons. Setting that equal to m*a and solving for a I get a=8.82m/s^2.

[EvoN1020v]

Ok, I refrained from looking at other posts for the first question, here's what I got:

 

 

There are 3 forces acting on the object: The normal force [math]F_N[/math], the weight vector [math]W[/math], and the frictional static [math]f_s[/math]. Since acceleration only occurs in the x direction, we only consider [math]\sum F_x=ma[/math]. To find [math]f_s[/math], which is the only force acting in the x direction, I convert the mass to kilograms, and multipliy the normal force ( [math]F_N=40.9*9.8=400.82[/math] ) by the frictional coefficient of .9 to get 360.738 Newtons. Setting that equal to m*a and solving for a I get a=8.82m/s^2.

 

Good job hotcommodity.

 

b) calculate the impulse.

 

Do you mind explaining more about that? To determine the velocities of the submarine and the torpedo, you don't need to calculate the impulse. Just use the momentum conversation theory and relative motion theory.

 

Anyone else tried this question? Or is it too hard for you guys?

[Klaynos]

Good job hotcommodity.

 

 

 

Do you mind explaining more about that? To determine the velocities of the submarine and the torpedo, you don't need to calculate the impulse. Just use the momentum conversation theory and relative motion theory.

 

Anyone else tried this question? Or is it too hard for you guys?

 

It wasn't something to calculate the first bit, it was an extra little problem, I can't even remember if it's possible with the data given as impulses are not something I use very often.

 

And lots of the people here are at least part way through an undergraduate degree in physics your question is not hard to them, and the people who could easily do it will have read it and decided to leave it for someone who likes a challenge.... That at least is what I did...

[EvoN1020v]

That's a lame excuse, Klaynos. Everyone is welcome to contribute and participate.

[Klaynos]

That's a lame excuse, Klaynos. Everyone is welcome to contribute and participate.

 

It's not lame it's the truth, your question is a 2 body conversation of momentum, I could do it but I don't want to put others who might enjoy it more from doing it.

 

It'd be like going to a primary school and answering all the teachers questions loudly from the back.

[EvoN1020v]

It's not lame it's the truth, your question is a 2 body conversation of momentum, I could do it but I don't want to put others who might enjoy it more from doing it.

 

It'd be like going to a primary school and answering all the teachers questions loudly from the back.

 

Then do you notice any students in the front answering the teacher's question?

 

[EvoN1020v]

Just a reminder: Three days left for the 2nd question before I put up another new question. (It will be easier, I promise you )

[dumbman29]

well the completely wrong answer i got to #2 was 4.98 m/s for the submarine and 12.78 or 8.89(both probably wrong) m/s for the torpedo

[EvoN1020v]

well the completely wrong answer i got to #2 was 4.98 m/s for the submarine and 12.78 or 8.89(both probably wrong) m/s for the torpedo

 

The velocity for the sumbmarine is 4.98 m/s which you got it right. ([math]V_s=4.981m/si - 0.00684m/sj[/math]). The speed of the torpedo is [math]V_T=3.413m/sj+14.38m/si[/math] or [math]14.78m/s[/math]. Your answer of 12.78 was close. Can you tell us how you got your calculations? That'd be appreciated.

 

Today is the day for a new question:

 

3rd question:

An electromagnetic launcher applies a force to 0.007 kg projectile that follows the illustrated force/time profile. How fast is the initially stationary mass moving at t=0.08s?

 

 

 

Good luck.

[dumbman29]

are you kidding lol? i didnt think either of my answers were even close to being right but the way i did it was:

50,000 x 5 = 250,000 (momentum of submarine)

100 x 10 = 1000 (mementum of torpedo)

250,000-1000=249,000

249,000/50,000=4.98 m/s

then for the torpedo i did:

7/9(10)+5=12.78

7/9 was there because there is 180 degrees that the torpedo could be launched. However 0-90 degrees and 90-180 degrees mimic the same effects on the velocitity of the torpedo. Therefore since the torpedo was facing 20 degrees up or down the velocity of the submarine would only be about 7/9 effective. I then added 5 for the original difference between the submarine's velocity and the torpedo's.

 

..I gotta learn how to right my work down when I do it, I had to redo the problem to find my work.

[EvoN1020v]

are you kidding lol? i didnt think either of my answers were even close to being right but the way i did it was:

50,000 x 5 = 250,000 (momentum of submarine)

100 x 10 = 1000 (mementum of torpedo)

250,000-1000=249,000

249,000/50,000=4.98 m/s

then for the torpedo i did:

7/9(10)+5=12.78

7/9 was there because there is 180 degrees that the torpedo could be launched. However 0-90 degrees and 90-180 degrees mimic the same effects on the velocitity of the torpedo. Therefore since the torpedo was facing 20 degrees up or down the velocity of the submarine would only be about 7/9 effective. I then added 5 for the original difference between the submarine's velocity and the torpedo's.

 

..I gotta learn how to right my work down when I do it, I had to redo the problem to find my work.

 

With the after collision of the momentum, you have to be careful. The mass of the submarine before the injection of the torpedo was 50,000 kg. Since the torpedo is 100 kg, the mass of the submarine will be 49,900 kg after the injection.

 

Try the 3rd question yourself. It's not that hard.

[EvoN1020v]

Anybody try the 3rd question? It's actually a high-school version type of question. Anyone should be able to answer it.