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Physics people! I need help!

p0ink

p0ink

New member
i only have one try left on this problem, and i need to make sure it's right.

this is easy stuff, and i know i am doing it right, but it says i'm not. so wtf?!

An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion.

F = m*g*cos(38)
sum forces parallel to the plane
m*g*sin(38) - (mu)*(F) = m*a, or
m*g*sin(38) - (mu)*m*g*cos(38) = m*a
masses cancel out
[9.81*sin(38)-(.281)]/[9.81*cos(38)] = .75 = mu

but it says i'm wrong. i cant be.
 
Considering parallel to plane...
(Force due to gravity) - (Force due to friction) = m * .281
Force due to friction = m * g * sin(38) - m * .281
F = 18.8 * (9.81 * 0.62 - .281)
F = 18.8 * (5.80)
F = 109.4 N

That's my stab at it. What is the known answer?
 
blut wump said:
Considering parallel to plane...
(Force due to gravity) - (Force due to friction) = m * .281
Force due to friction = m * g * sin(38) - m * .281
F = 18.8 * (9.81 * 0.62 - .281)
F = 18.8 * (5.80)
F = 109.4 N

That's my stab at it. What is the known answer?
Click to expand...

i dont know the answer.

it's the frictional force that it's aking for, right? i know my mu is correct...but it says it isnt.
 
Our calculations are similar except that you've divided by the gravity force into the plane. Then again, maybe you're looking for the coefficient of friction rather than the actual force resisting the motion.
 
and this one:

Tank A

|..........|
|..h2o...|
|..h2o...|
|_h2o__|

Tank B

\.........../
.\..h2o../
..\.h2o./
...\h2o/

Tank C

.../.....\
../.h2o.\
./..h2o..\
/__h2o__\


The three tanks shown above are filled with water to an equal depth. All the tanks have an equal height. Tank A has the middle surface area at the bottom, tank B the least and tank C the greatest.

(Select T-True, F-False, G-Greater than, L-Less than, E-Equal to. If the first is F the second L and the rest G, enter
FLGGGG).

A) The force due to the water on the flat bottom of tank B is .... the weight of the water in the tank.
B) The force exerted by the water on the bottom of tank C is ... the force exerted by the water on the bottom of tank B
C) The water in tank B exerts a force with an upward component on the sides of the tank.
D) The pressure exerted on the bottom of tank C is greater than for the other tanks.
E) The pressure at the bottom of tank C is ... the pressure at the bottom of tank B
 
ok, i know this much

A: F=mg
B: F=mg-(the upward force of the sides)
C: F=mg+(the downward force of the sides)

the wording is throwing me off here.
 
For the first one:

define the positive x axis as being on the plane that the object is sliding on.

The sum of the forces in x:

0 = -F(friction) + mg*sin(theta) + m(given acceleration)

F(friction) = m(a + gsin(theta))
 
UA_Iron said:
For the first one:

define the positive x axis as being on the plane that the object is sliding on.

The sum of the forces in x:

0 = -F(friction) + mg*sin(theta) + m(given acceleration)

F(friction) = m(a + gsin(theta))
Click to expand...


nope, that isn't it. i know your thinking here, but that's not correct.

it's asking for the coefficient of friction, correct? so why isn't my mu = .75?
 
well in the original question you asked for the magnitude of the friction force, not the coefficient of friction.

mu = (gsin(theta) + a) / (gcos(theta))
 
UA_Iron said:
well in the original question you asked for the magnitude of the friction force, not the coefficient of friction.

mu = (gsin(theta) + a) / (gcos(theta))
Click to expand...

blah, dont mind me....it's been a trying day

what numerical answer are you getting....i dont trust my ability to remember my way home, let alone my math skills.

plus, i am doing this all on my cell phone's calculator.
 
blut wump said:
Considering parallel to plane...
(Force due to gravity) - (Force due to friction) = m * .281
Force due to friction = m * g * sin(38) - m * .281
F = 18.8 * (9.81 * 0.62 - .281)
F = 18.8 * (5.80)
F = 109.4 N

That's my stab at it. What is the known answer?
Click to expand...

At a glance, this is the credited response. But I am beginning to believe Samoth when we need laTex enabled text.
 
I havent had a physics class for like 3 years (college physics 1&2) but I did have Telecommunication electronics this year but that wont help.

This is making my head hurt!
 
redguru said:
At a glance, this is the credited response. But I am beginning to believe Samoth when we need laTex enabled text.
Click to expand...

From what I recall on physicsforums, it's somewhat complicated and expensive to implicate... however, I'm not sure about that. I just remember it was a big exciting deal when it was finally integrated into the forums.

Unfortunetly, no one here would have the patience to learn how to use it... :rolleyes:



:cow:
 
samoth said:
From what I recall on physicsforums, it's somewhat complicated and expensive to implicate... however, I'm not sure about that. I just remember it was a big exciting deal when it was finally integrated into the forums.

Unfortunetly, no one here would have the patience to learn how to use it... :rolleyes:

:cow:
Click to expand...

have the patience to use what?

what do you think samoth? i only have one chance left.
 
blut wump said:
Considering parallel to plane...
(Force due to gravity) - (Force due to friction) = m * .281
Force due to friction = m * g * sin(38) - m * .281
F = 18.8 * (9.81 * 0.62 - .281)
F = 18.8 * (5.80)
F = 109.4 N
Click to expand...

Once again I agree with him. Go with this.
 
3 + 5 = 8

Can someone confirm this?

I'd get pretty excited just to spell physics correctly, let alone do it. I keep it simple.
 
Indeed, I was happier to approximate than you might be. I assumed you'd take your own desired accuracy when invoking the method. My bad, I guess.

F = 18.8 * (9.81 * sin(38) - 0.281)
F = 108.262

Be sure that you're looking for forces rather than coeff. of friction, though.
 
p0ink said:
i only have one try left on this problem, and i need to make sure it's right.

this is easy stuff, and i know i am doing it right, but it says i'm not. so wtf?!

An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion.

F = m*g*cos(38)
sum forces parallel to the plane
m*g*sin(38) - (mu)*(F) = m*a, or
m*g*sin(38) - (mu)*m*g*cos(38) = m*a
masses cancel out
[9.81*sin(38)-(.281)]/[9.81*cos(38)] = .75 = mu

but it says i'm wrong. i cant be.
Click to expand...
OMG, why the crap did I even read this thread? This brought back bad, bad memories of physics during college. To this day I still don't see why it's a prerequisite for medical school. Just the other day I was intubating and placing a central line in a septic patient and thought to myself, hmm, I wonder what force is required to insert this triple lumen catheter without dilating v. dilating? How many newtons is that again?
 
swatdoc said:
OMG, why the crap did I even read this thread? This brought back bad, bad memories of physics during college. To this day I still don't see why it's a prerequisite for medical school. Just the other day I was intubating and placing a central line in a septic patient and thought to myself, hmm, I wonder what force is required to insert this triple lumen catheter without dilating v. dilating? How many newtons is that again?
Click to expand...

are you a swat paramedic? how do you enjoy that? my buddy has been one for quite awhile now.
 
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