## Trending News

# 物理題求解

A 5.0-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?

Two blocks with masses m1=6kg and m2=2kg are connected by a string that hangs

over a pulley of mass 4kg and radius 10cm.

(a) What is the net torque on the system?

(b) What is the angular momentum on the system when the blocks have the speed V(c) Find the acceleration of the blocks by applying the equation t = dL/dt.

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? (b) the center of mass?

(c) the moment of inertia for rotation about the end of Y= 1kg/m?

前2題的圖 http://imgur.com/XKRFcan

做了一些題目 這3題是看了答案以後還是不會寫的 求詳解~~感謝

### 2 Answers

- 麻辣Lv 76 years agoFavorite Answer
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)