1.A 5 m long ladder leans against a smooth wall at a point 4.0 m above
a cement floor. The ladder is uniform and has mass m=10kg. Assuming the
wall is frictionless, but us=0.5 for the floor. What is the maximum
distance along the ladder a person of mass 50kg can climb before the
ladder start to slip?
N1=(m+M)g=60g
f=us*N1=0.5N1=30g
N2=f=30g
ΣM(ladder bottom)=Mg*x+mg*3/2-N2*4=0
x=(4N2-3mg/2)/Mg
=(4*30g-30g/2)/50g
=(120-15)/50
=105/50
=2.1 (m)
L=x/sin37
=2.1/0.6
=3.5 (m)...ans
2.Two blocks with masses m1=6kg and m2=2kg are connected by a string
that hangs over a pulley of mass m=4kg and radius r=10cm.
(a) What is the net torque on the system?
T=(m1-m2)g*r
=(6-2)*g*10
=40g
=392 (N.m)
(b) What is the angular momentum on the system when the blocks have the
speed V?
I=Momentum of Inertia
=0.5mr^2
=0.5*4*10^2
=200 (kg.m^2)
w=Angular speed
=V/r
=V/10 (rad/s)
M=Angular momentum
=I*w
=200*V/10
=20V
(c) Find the acceleration of the blocks by applying the equation
t=dL/dt.(The Eq. is error)
α=T/I
=392/200
=1.96 (rad/s^2)
a=rα
=10*1.96
=19.6 (m/s)
3.A rod with length L=1.8m, but its linear mass density Y varies
linearly from Y=1kg/m Y at the left end to double that value, Y = 2
kg/m, at the right end. Find
(a) the mass of this rod?
y = dm/dx => dm = ydx
Left = (0,1)
Right = (1.8,2)
Linear equation: (y-1)/x = (2-1)/1.8
=> y(x) = 1 + x/1.8 =>
m = ∫y(x)dx
= ∫(1+x/1.8)dx ... x=0~1.8
= x + x^2/3.6
= 1.8 + 1.8*1.8/3.6
= 1.8 + 0.9
= 2.7 (kg)
(b) the center of mass?
xbar*mass=∫x*dm
=∫xy*dx ... dm=ydx
=∫x(1+x/1.8)dx
=∫(x+x^2/1.8)dx
= x^2/2 + x^3/5.4 ... x=0~1.8
= 1.8^2/2 + 1.8^3/5.4
= 2.7
xbar = 2.7/2.7 = 1 (m)
(c) the moment of inertia for rotation about the end of Y= 1kg/m?
I=∫x^2*dm
=∫x^2*ydx
=∫x^2*(1 + x/1.8)dx
=∫(x^2 + x^3/1.8)dx
= x^3/3 + x^4/7.2
= x^3*(1/3 + x/7.2)
=1.8^3*(1/3 + 1.8/7.2)
=(1/3 + 1/4)1.8^3
=7*1.8^3/12
=3.402 (m^4)
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