yy205001
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yy205001 said:Homework Statement
What is the acceleration of the 2. kg block in the figure across the frictionless table?
Homework Equations
F=ma
The Attempt at a Solution
g = 9.8ms^2
yy205001 said:yea, i drew FBD for both blocks. In the x-direction, the resulting force is T. And in the y direction is (m(1kg)g-T).
gneill 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?
gneill said: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:gneill, do you mean the acceleration of the hanging mass is twice the acceleration of the mass on the horizontal surface??
Tension has units of Newtons (force). It is not an acceleration (m/s2).yy205001 said: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
Again you're mixing units. Forces (like m*g) are not accelerations (m/s2).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?