- #1
How can i find the acceleration for this constraint?
- Thread starter yy205001
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To find the acceleration of a 2 kg block on a frictionless table connected to a hanging mass, it's essential to analyze the forces acting on both blocks using free body diagrams (FBDs). The tension in the rope and the gravitational force on the hanging mass create a constraint that relates their accelerations. The acceleration of the hanging mass is twice that of the block on the table due to the pulley system's geometry. By applying Newton's second law and substituting the relationship between the accelerations, the correct acceleration can be derived. Ultimately, understanding the constraints imposed by the system is crucial for solving the problem accurately.
- #2
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As we welcome you to these forums, you first should show more of an attempt before we can assist. You need to draw force diagrams of each block and apply Newton's laws. Do you think the acceleration of the blocks are the same?
- #3
yy205001
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as long as they are in the same system, i think they have the same acceleration
- #4
gneill
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yy205001 said:
You'll have to show more effort. What is the constraint imposed by the pulley system on the relative motions of the blocks? Did you draw FBD's for both blocks? What are the resulting force equations?
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gneill
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yy205001 said:
Okay. In your diagram you depict two "T" forces acting on the 2kg block, so why do you say the resulting force is T? How are the accelerations of the blocks related?
- #7
yy205001
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gneill said:
sorry, it should be 2T. Are they having the same acceleration, this is my main problem.
- #8
gneill
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The rope and pulley system place a constraint on the motions. If the 1kg block moves down a distance d, how far will the 2kg block slide?
- #9
yy205001
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will it slides a distance d as well??
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- #10
gneill
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yy205001 said:
Nope. Do a rope segment length calculation. Remember that the rope's total length is constant.
- #11
yy205001
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yea,i got d/2. Is it correct??
- #12
gneill
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Yes, that's fine. That then will be the constraint connecting the motions of the blocks. It will apply to distance moved, velocities, and accelerations.yy205001 said:
- #13
yy205001
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gneill said:
i still cannot get it
- #14
yy205001
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gneill, do you mean the acceleration of the hanging mass is twice the acceleration of the mass on the horizontal surface??
- #15
gneill
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yy205001 said:
Yes, it must be due to the constraint...
Convert your FBD force equations to acceleration for each block. Use the constraint equation for acceleration to solve.
- #16
yy205001
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In the x-direction, the Net force is 2*T.
Let a1 be the acceleration of the mass on horizontal surface,
a2 be the acceleration of the hanging mass.
"F=m*a", so:
2*T = 2*a1
T = a1
In the y-direction, the Net force is mg-T, where m is 1kg,so:
m*g-T = m*a2
where T = a1
→m*g-a1=m*a2
Then,sub in a2=2*a1
m*g-a1=m*2*a1
So, i can find out the acceleration of a1, thus find a2?
Let a1 be the acceleration of the mass on horizontal surface,
a2 be the acceleration of the hanging mass.
"F=m*a", so:
2*T = 2*a1
T = a1
In the y-direction, the Net force is mg-T, where m is 1kg,so:
m*g-T = m*a2
where T = a1
→m*g-a1=m*a2
Then,sub in a2=2*a1
m*g-a1=m*2*a1
So, i can find out the acceleration of a1, thus find a2?
- #17
gneill
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Tension has units of Newtons (force). It is not an acceleration (m/s2).yy205001 said:
Use the relationship F = ma that you've written to turn the net force into an acceleration, hence a = F/m. So that would make a1 = ?
Again you're mixing units. Forces (like m*g) are not accelerations (m/s2).
Just write the accelerations for each block separately at first; Don't try to apply the constraint until you have those expressions. You should have:
a1 = (something in terms of m1 and T)
a2 = (something in terms of m2 and T and g)
After you have them you can use the constraint equation to eliminate one of the accelerations...
- #18
yy205001
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yes, i got the correct answer finally, thank you so much!