物理題求解

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=60gf=us*N1=0.5N1=30gN2=f=30gΣM(ladder bottom)=Mg*x+mg*3/2-N2*4=0x=(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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