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mc8569
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Newton's Law of Motion problem! Help please!
A small block of mass m1 is placed on block 2 of mass m2 this is then placed on the table. A string connecting block 2 to a hanging mass M passes over a pulley attached to one end of the table. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and 2 and block one and the tabletop are nonzero and are:
Coeff. between 1 & 2:
Static = (mus1) <-- mu sub static 1
Kinetic = (muk1) <-- mu sub kinetic 1
Coeff. between 2 & Table:
Static = (mus2)
Kinetic = (muk2)
c) M is large enough that the hanging blocks descend when the block is released. Assume that blocks 1 and 2 are moving as a unit (no slippage) Determine the magnitude a of the acceleration.
d) Now suppose that M is large enough that as the hanging block desends, block 1 is slipping on block 2
Determine the following:
i. The magnitude a1 of the acceleration of block 1
ii. The magnitude a2 of the acceleration of block 2
For c) I got:
a = g[M - (muk2)(m1 + m2)] / (m1 + m2 + M)
^^after factoring out g, correct?
Real concern is with d)
d) i. F = ma
fk1 = m1a1 = (muk1)n1
a1 = (muk1)m1g / m1
*m1 cancels out
a1 = (muk1)g
Is that right?
ii. was what confused me,
Should (ii) be the same answer as part (c)? Please help me on how to approach this if not. I am thinking that the reverse movement of m1 on m2 would decrease the acceleration, but I thought, as a system it shouldn't do anything because the friction on m1 is equal and opposite to m2. It isn't an external force. Help please?
Homework Statement
A small block of mass m1 is placed on block 2 of mass m2 this is then placed on the table. A string connecting block 2 to a hanging mass M passes over a pulley attached to one end of the table. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and 2 and block one and the tabletop are nonzero and are:
Coeff. between 1 & 2:
Static = (mus1) <-- mu sub static 1
Kinetic = (muk1) <-- mu sub kinetic 1
Coeff. between 2 & Table:
Static = (mus2)
Kinetic = (muk2)
Homework Equations
c) M is large enough that the hanging blocks descend when the block is released. Assume that blocks 1 and 2 are moving as a unit (no slippage) Determine the magnitude a of the acceleration.
d) Now suppose that M is large enough that as the hanging block desends, block 1 is slipping on block 2
Determine the following:
i. The magnitude a1 of the acceleration of block 1
ii. The magnitude a2 of the acceleration of block 2
The Attempt at a Solution
For c) I got:
a = g[M - (muk2)(m1 + m2)] / (m1 + m2 + M)
^^after factoring out g, correct?
Real concern is with d)
d) i. F = ma
fk1 = m1a1 = (muk1)n1
a1 = (muk1)m1g / m1
*m1 cancels out
a1 = (muk1)g
Is that right?
ii. was what confused me,
Should (ii) be the same answer as part (c)? Please help me on how to approach this if not. I am thinking that the reverse movement of m1 on m2 would decrease the acceleration, but I thought, as a system it shouldn't do anything because the friction on m1 is equal and opposite to m2. It isn't an external force. Help please?