## A Couple Seemingly Simple Physics Problems

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codyhotel
Posts: 319
Joined: Mon Dec 03, 2007 3:54 pm UTC

### A Couple Seemingly Simple Physics Problems

I have an exam tomorrow, so time is of the essence, but if not done in time I'm taking a successive course in January so still need the answers.

1.

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A uniform ladder of length L and mass m rests against a wall at some angle. There is no friction between the ladder and the wall. If the coefficient of static friction between the ladder and floor is 0.2, what is the smallest angle \theta such that the ladder won't slip? (The floor and wall make a right angle, \theta is the angle between the floor and the ladder)

Spoiler:
My solution is to set \Mu_s=f_s/F_n=[mgLcos\theta]/[mgLsin\theta]=cot\theta, and then tan^-1(1/.2)=\theta=78.7 degrees. but the answer key says 68.2 degrees, which I can't figure out.

2.

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A block of iron of volume 10^-5m^3 rests on the bottom of a tank of fluid and exerts a force of 0.500N on the bottom of the tank. What is the density of the fluid if the density of Iron is 7.87x10^3kg/m^3?

Spoiler:
I have no idea how to do this to be honest

Thanks in advance for any help guys.
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eta oin shrdlu
Posts: 451
Joined: Sat Jan 19, 2008 4:25 am UTC

### Re: A Couple Seemingly Simple Physics Problems

For the first problem, you are off by a factor of 2. Check your free-body diagram and make sure you have the forces applied at the correct locations.

For the second problem, the forces acting on the iron block are: a downward force due to gravity, an upward buoyant force, and the supporting force from the bottom of the tank. What is the magnitude of the buoyant force?

Rgeminas
Posts: 45
Joined: Mon Aug 11, 2008 5:02 pm UTC

### Re: A Couple Seemingly Simple Physics Problems

For the second problem, remember that the the sum of the forces in the vertical direction must be zero, since it is stationary. Remember how the buoyant force equals the weight of the fluid dislodged by the cube.

Solving for a general case:
Spoiler:
Solving for a general case: There is a body of volume [imath]V_1[/imath], density [imath]\rho_1[/imath], sitting in a tank with a fluid, exerting a force on the floor F. Find the density of the fluid (which I'll call [imath]\rho_2[/imath]).

There are three forces: buoyancy, pointing up and measuring [imath]\rho_2.V_1.g[/imath], where g is the intensity of the gravitational field, the body's weight, pointing down and measuring [imath]\rho_1.V_1.g[/imath](which is the same as m.g, for [imath]m=\rho.V[/imath], and the contact force, which is F, pointing up as well.

As the body is not accelerating, [imath]\rho_2.V_1.g+F-\rho_1.V_1.g=0[/imath]. Since you have both densities, [imath]V_1[/imath], g and F, solve for [imath]\rho_2[/imath].The result would be: [imath]\rho_2=\frac {\rho_1.V_1.g-F} {V_1.g}[/imath].

For the first problem, my answer was the same as yours, so sorry.

danpilon54
Posts: 322
Joined: Fri Jul 20, 2007 12:10 am UTC

### Re: A Couple Seemingly Simple Physics Problems

for the first one, where along the ladder can you say the force is applied? I believe you used that its applied to the top. *hint*
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codyhotel
Posts: 319
Joined: Mon Dec 03, 2007 3:54 pm UTC

### Re: A Couple Seemingly Simple Physics Problems

Okay, I see where I went wrong on the free body diagram for the first question and got the right answer, thanks guys.

For the second problem I think there must be a mistake in the answer key, Rgeminas I know your solution is right but I still don't get the right answer, but if I use a Force of 5N rather than 0.5N, I do get the right answer. Unless there is something I'm not getting here.
Philwelch wrote:Would a prostitution enthusiast be a buy-sexual?

...sorry.

Twinfire0
Posts: 23
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Location: United States

### Re: A Couple Seemingly Simple Physics Problems

codyhotel wrote:Okay, I see where I went wrong on the free body diagram for the first question and got the right answer, thanks guys.

For the second problem I think there must be a mistake in the answer key, Rgeminas I know your solution is right but I still don't get the right answer, but if I use a Force of 5N rather than 0.5N, I do get the right answer. Unless there is something I'm not getting here.

In helping friends with similar physics problems in the past (we actually just got finished covering the basics of fluid physics), I've found that in most calculations involving masses and forces where the final answer is off by a factor of around 10, perhaps you are forgetting the acceleration of gravity somewhere in your calculations? I'll take a look at the problem, and re-edit this post once I get an answer.

2.

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A block of iron of volume 10^-5m^3 rests on the bottom of a tank of fluid and exerts a force of 0.500N on the bottom of the tank. What is the density of the fluid if the density of Iron is 7.87x10^3kg/m^3?

Solution:
Spoiler:
Okay, so first we have to find the mass of the block of iron (use density*volume=mass), which is equal to .0787 kg. Multiply by the force of gravity and you get .77126 N. Subtract the measured force exerted by the object (.5 N) and you'll get its buoyant force, which is .271 N. Divide by the acceleration of gravity to get the mass of the fluid which is displaced by the block of iron, which is .0277 kg. Now, finally divide this mass by the volume of fluid displaced and you'll get a density of the fluid, which turns out to be 2.77*10^3 kg/m^3.

Hope that helps!

codyhotel
Posts: 319
Joined: Mon Dec 03, 2007 3:54 pm UTC

### Re: A Couple Seemingly Simple Physics Problems

I didn't actually forget anything, I just did the calculations wrong giving me a mass of .787kg instead of .0787kg, I must have entered that 10^-5 wrong. Thanks!

Luckily there was no question like that on the exam! (In my opinion though, physics exams should be all application questions, none of this theory garbage, tryin to screw me up,...makin me use mah logic skillz,....)
Philwelch wrote:Would a prostitution enthusiast be a buy-sexual?

...sorry.

roundedge
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