Mass mA rests on a smooth horizontal surface, mB hangs vertically.If m_{\rm{A}}=13.0 kg and m_{\rm{B}}=3.0 kg in the figure , determine the magnitude of the acceleration of each block.
you smell like celery and gulumpkies
The figure shows a block (mass massA ) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (mass
To determine the magnitude of acceleration of each block, we need to consider the forces acting on them.
For block A (mA):
1. The only force acting on block A is the tension force (T) from the string connecting it to block B.
2. Since the surface is smooth, there is no frictional force.
3. Using Newton's second law (F = ma), we have:
T = mA * aA
where T is the tension force and aA is the acceleration of block A.
For block B (mB):
1. There are two forces acting on block B: the weight force (mg) acting downward and the tension force (T) from the string acting upward.
2. Using Newton's second law (F = ma), we have:
mg - T = mB * aB
where mg is the weight force, T is the tension force, and aB is the acceleration of block B.
To find the magnitude of acceleration for each block, we can solve these two equations simultaneously.
From the first equation, T = mA * aA. Substituting this into the second equation, we have:
mg - mA * aA = mB * aB
Now, substitute the given values:
mB = 3.0 kg
mA = 13.0 kg
g = 9.8 m/s^2 (acceleration due to gravity)
Plugging in these values, we have:
(3.0 kg) * (9.8 m/s^2) - (13.0 kg) * aA = (3.0 kg) * aB
Simplifying this equation gives:
29.4 - 13.0 * aA = 3.0 * aB
From the first equation, T = 13.0 kg * aA. Substituting this into the second equation, we have:
29.4 - 13.0 * (T / 13.0 kg) = 3.0 * aB
Simplifying this equation gives:
29.4 - T = 3.0 * aB
Now, substitute T = 13.0 kg * aA in this equation:
29.4 - 13.0 * (13.0 kg * aA) = 3.0 * aB
Simplifying this equation gives:
29.4 - 169.0 * aA = 3.0 * aB
Solving these equations simultaneously will give us the values of aA and aB. Since the equations are non-linear, substitution or substitution methods can be used for solving the equations.
Sorry, Anonymous, we do not have latex installed, at least yet.
I presume the figure looks like this, but a mirror image of it:
o-----T-----mA
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T
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mB
The letter o stands for a pulley.
The answer to the question lies in the formula F=ma
Force exerted on the system is gravity on mB, thus
F=mB*g
Total mass = mA+mB
Acceleration = F/(mA+mB)
If they are connected by a mass-less and frictionless pulley, then the acceleration of each block is identical.
The tension on the string, T, is again found by the formula:
T=mA*a