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> showed my work please help
helokit
Posted: Feb 20 2006, 07:14 PM


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The bolts on the cylinder head of certain engines require tightening to a torque of 86 mN. If a wrench is 26 cm long, what force perpendicular to the wrench must the mechanic exert at its end?
86=.26N
N=3.30e2N

If the six-sided bolt head is 15 mm in diameter, estimate the force applied near each of the six points by a socket wrench.
I don't know how to do this one. I got 90=.045*6*N
N=1911.1N
I got .045m from converting 15mm to meters divided it by half to get radius. Can you please tell me what i did wrong and explain it to me

There's a picture of a wrench with two hexagons at the end. The left side one has force downward and the diameter is 15mm. On the right side, there is force downward on the wrench.

How do i post a picture because my homework is online. There's another post whwere somebody posted a picture.


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helokit
Posted: Feb 20 2006, 07:48 PM


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A centrifuge rotor rotating at 6000 rpm is shut off and is eventually brought to rest by a frictional torque of 1.50 mN. If the mass of the rotor is 4.70 kg and it can be approximated as a solid cylinder of radius 0.0780 m, through how many revolutions will the rotor turn before coming to rest?
rev
How long will it take?
s


I conveted rpm to rev/s which is 100rev/s and i don't what to do next. How do bring the rev/s into the equation. please help me. i have a test tomorrow. What do i do with the mass.
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Moseley
Posted: Feb 21 2006, 01:13 AM


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Inertia is the word you are looking for. You will have a formula somewhere for this - when you find out what inertia the rotor has divide by stopping force to find time.
It is something like J=1/2*m*r*w^2 where w is angular velocity in radians/sec and r is radius.
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WebAssign
Posted: Mar 15 2012, 11:12 PM


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Disclaimer: I know this is 6 years too late, but this is to help current sufferers of webassign and similar online homework:

we're going to start with part B and then come back to part A:

b) delta time
for a solid cylinder, I = .5*M*r^2
I = .5 * 4.7kg * (0.078m)^2
I = 0.0143 kgm^2

now, based on the equations J = delta J / delta t J = I * w
you can rearrange to get T = (I * initial w) / delta t
change your initial w from rpm to radians/ sec:
the initial w = 6000 rpm*(2m/rev)*(min/60 sec) ==> 628 radians/sec
so...
delta t = (I * initial w) / T
delta t = (0.0143 kgm^2 * 628 radians/sec) / 1.5 mN
delta t = 5.99 seconds

now, back to part a) (revolutions before stopping)

kinematics!
initial v (let's call that v0) = 6000 rpm
we just want to adjust the time component this time, so 6000rev/min * min/60 sec ==> 100 rev/ sec
vf = v0 + a * t
0 rev/sec = 100 rev/sec + a * (5.99 sec)
a = -16.69 rev/ sec^2

now use equation:
delta x = (v0 * t) + (.5 * a * t^2)
delta x = (100 rev/sec * 5.99 sec) + (.5 * -16.69 rev/sec^2 * 5.99 sec^2)
delta x = 299.5 revolutions





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