fisica1-youn-e-freedman-exercicios-resolvidos

5.121: If the block is not to move vertically, the acceleration must be horizontal. The
common acceleration is a = g tan θ,
so the applied force must be
( M + m)
a = ( M + m)
g tan θ.
5.122: The normal force that the ramp exerts on the box will be n = w cos α − T sin θ.
The rope provides a force of T cos θ up the ramp, and the component of the weight down
the ramp is w sin α.
Thus, the net force up the ramp is
F = T cos θ − wsin
α − µ ( w cos α − T sin θ)
= T (cos θ + µ
k
sin θ)
− w(sin
α + µ
k
cos α).
The acceleration will be the greatest when the first term in parantheses is greatest; as in
Problems 5.77 and 5.123, this occurs when tan θ = µ .
k
k
5.123: a) See Exercise 5.38; F = µ
k
w (cos θ + µ
k
sin θ).
b)
c) The expression for F is a minimum when the denominator is a maximum; the
calculus is identical to that of Problem 5.77 (maximizing w for a given F gives the same
result as minimizing F for a given w), and so F is minimized at tan θ = µ
k.
For
µ = 0.25, θ = 14.0 , keeping an extra figure.
k
°