- #1
weinbergshaun
- 12
- 1
Question:
A large box of mass M is pulled across a horizontal, friction-less surface by a horizontal rope with tension T. A small box mass m sits on top of the box. The coefficients of static and kinetic frciton between the two boxes are Us and Uk. find an expreeion for the maximum tension Tmax for which the small box rides on top of the large box without slipping. Relevant equations
None given
Know formulas:
FFriction (or FF) = m2 x a (the friction of the top on lower box)
FTension (or FT) - FFriction = m1 x a
the two forces must have the same acceleration in the second formula, will get both acc. alone on both formulasand equal them.
Top block acc. is:
FT = m2 x a (get a alone)
a = FF/m2
Bottom block acc. is:
FT - FF = = m1 x a
FT – (m2 x a) = m1 x a (substituting Ff for the m2 x a value in order to have one force value only)
FT = m1 x a + m2 x a
FT = a(m1 + m2)
a = FT/(m1 + m2)
a = a, getting Tension or FT alone:
FF/m2 = FT/(m1 + m2)
FT = FF(m1 + m2)/m2
FMax = m1 x g x Us (m1 + m2)/m2 (substituting FT with max as it will represent max).
Is this correct, thank
A large box of mass M is pulled across a horizontal, friction-less surface by a horizontal rope with tension T. A small box mass m sits on top of the box. The coefficients of static and kinetic frciton between the two boxes are Us and Uk. find an expreeion for the maximum tension Tmax for which the small box rides on top of the large box without slipping. Relevant equations
None given
The Attempt at a Solution
Know formulas:
FFriction (or FF) = m2 x a (the friction of the top on lower box)
FTension (or FT) - FFriction = m1 x a
the two forces must have the same acceleration in the second formula, will get both acc. alone on both formulasand equal them.
Top block acc. is:
FT = m2 x a (get a alone)
a = FF/m2
Bottom block acc. is:
FT - FF = = m1 x a
FT – (m2 x a) = m1 x a (substituting Ff for the m2 x a value in order to have one force value only)
FT = m1 x a + m2 x a
FT = a(m1 + m2)
a = FT/(m1 + m2)
a = a, getting Tension or FT alone:
FF/m2 = FT/(m1 + m2)
FT = FF(m1 + m2)/m2
FMax = m1 x g x Us (m1 + m2)/m2 (substituting FT with max as it will represent max).
Is this correct, thank