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May 16th, 2017, 10:52 PM   #1
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Need help - Newton's second law of motion

A 2.0 ton elevator is supported by a cable that can safely support 6400 lb. What is the shortest distance in which the elevator can be brought to a stop when it is descending with a speed of 4.0 ft/s?


A 2.0 ton elevator is supported by a cable that can safely support 6400 lb. What is the shortest distance in which the elevator can be brought to a stop when it is descending with a speed of 4.0 ft/s?


A 50-kg block is placed on a smooth horizontal surface. A horizontal chord attached to the block passes over a light frictionless pulley and is attached to a 4.0-kg body. Find the acceleration and the tension in the chord when the system is released.


Two bodies having the masses m1=30g and m2=40g are attached to the ends of a string of negligible mass and suspended from a light frictionless pulley. Find the accelerations of the bodies and the tension in the string.
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May 17th, 2017, 03:51 AM   #2
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Well, you posted the first problem twice, so I suppose it's important to start there.

Maximum upward net force would be $6400 - 4000 = 2400 \, lbs$. Using $g= 32 \, ft/s^2$ ...

$a_{max}=\dfrac{F_{net}}{m}=\dfrac{2400}{125} = 19.2 \, ft/s^2$

You may now use the kinematics equation, $v_f^2 = v_0^2 + 2a \Delta x$, to determine the minimum displacement required to bring the elevator to a stop.



For the second problem, I assume you've sketched a diagram.

Net force on the table mass is just tension ...

$T = Ma$

Net force on the hanging mass is its weight minus the same tension ...

$mg - T = ma$

Solve the system of equations for acceleration, $a$, then determine the tension, $T$.


Atwood machine problem ... net force on each mass again.

$m_2 g - T = m_2 a$

$T - m_1 g = m_1 a$

summing the two equations eliminates $T$ ...

$m_2 g - m_1 g = m_2 a + m_1 a$

You may now solve for the magnitude of acceleration, $a$, then go back and determine the value of tension, $T$.
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May 17th, 2017, 04:24 AM   #3
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Thank you so much. Great help!
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May 17th, 2017, 04:36 AM   #4
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Where did you get the 6400-4000=2400 lbs?
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May 17th, 2017, 05:32 AM   #5
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see attached force diagram ... like the one you made, correct?
Attached Images
File Type: jpg Fnet.jpg (20.9 KB, 5 views)
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May 17th, 2017, 07:00 AM   #6
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Thanks! How about this problem?

Find the torque created by a 10-N force acting 60°N of E at a distance of 1m from the axis of rotation of a lever.
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May 17th, 2017, 07:16 AM   #7
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Quote:
Originally Posted by Camsybel View Post
Thanks! How about this problem?

Find the torque created by a 10-N force acting 60°N of E at a distance of 1m from the axis of rotation of a lever.
$\tau = \vec{r} \times \vec{F} = |\vec{r}| \, |\vec{F}| \, \sin{\theta}$

you have been given the magnitude & direction for $\vec{F}$, but only the magnitude of $\vec{r}$ ... what is the direction of $\vec{r}$? Is there a diagram given?
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May 17th, 2017, 07:34 AM   #8
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There is no diagram given.
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May 17th, 2017, 07:53 AM   #9
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Quote:
Originally Posted by Camsybel View Post
There is no diagram given.
I hate to assume ... maybe the statement infers that $\theta = 60^\circ$

Use the torque formula I provided in my previous post.

Poorly worded question, imho.
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