What are the full solutions of some mechanics questions:
Q1: A light elastic string of natural length 0.2m has its ends attached to 2 fixed points A and B which are on the same horizontal level with AB=0.2m. A particle of mass 5kg is attached to the string at the point P where AP=0.15m. The system is released and P hangs in equilibrium below AB with [angle APB= 90 degrees].
a) If angle BAP = x, show that the ratio of the extension of AP and BP is [(4cosx - 3)/(4sinx - 1)]
b) Hence show that (cosx)(4cosx - 3)=(3sinx)(4sinx - 1)
Q2: A particle of mass 3kg is attached to one end of a light string, of natural length 1m and modulus of elasticity 14.7N. The other end of the string is attached to a fixed point. The particle is held in equilibrium by a horizontal force of magnitude 9.8N with the string inclined to the vertical at an angle x.
If the horizontal force is removed, find the magnitude of the least force that will keep the string inclined at the same angle.
Q3: The particle P of mass 4kg is attached to one end of a light inelastic string of length 1m. The other end of the string is fixed at point O. P is hanging in equilibrium below O when it is projected horizontally with speed u ms-1. When OP is horizontal it meets a small smooth peg at Q, where OQ=0.8m. Calculate the minimum value of u if P is to describe a complete circle about Q.
Mechanics
2013-05-31 1:42 am
回答 (1)
2013-06-03 11:46 pm
✔ 最佳答案
1. Let Ta be the tension of AP and Tb be the tension of BPSince ∠BAP=x and ∠APB=90°, ∠ABP=90°-x
cos∠ABP=sinx, sin∠ABP=cosx
In equilibrium, Ta*cosx-Tb*sinx=0,
Ta/Tb=tan(x)
AP=0.15m, BP=0.2-0.15=0.05m=AP/3
Since the original length of AP is 3 times that of BP and they have the same spring constant,
Extension of AP = 3*Extension of BP
The ratio of the extension of AP and BP is 3Ta/Tb=3tanx
Let C be a point on A such that AC=0.15m, then CP=Extension of AP
Let D be a point on B such that BD=0.05m, then DP=Extension of BP
Then In ABP, AP^2+BP^2=(0.15+CP)^2+(0.05+DP)^2=AB^2=0.2^2
In △ABD, AD^2=0.2^2+0.05^2-2*0.2*0.05*cos(90°-x)=0.0425-0.02sinx
In △ADP, AD^2=AP^2+DP^2=(0.15+CP)^2+DP^2
AP^2+(0.05+DP)^2-[AP^2+DP^2] = 0.04-[0.0425-0.02sinx]
0.0025+0.1DP=-0.0025+0.02sinx
DP = 0.2sinx-0.05
BC^2 = CP^2+BP^2 = 0.2^2+0.15^2-2*0.2*0.15cosx = 0.0625-0.06cosx
(0.15+CP)^2+(0.05+DP)^2-[CP^2+(0.05+DP)^2]
=0.0025+0.3CP = 0.04-[0.0625-0.06cosx]
CP = 0.2cosx+0.15
Ratio of the extension of AP and BP
=CP/DP
=[0.2cosx+0.15]/[0.2sinx-0.05]
=(4cosx+3)/(4sinx-1)
Therefore 3tanx=(4cosx+3)/(4sinx-1)
(cosx)(4cosx+3)=(3sinx)(4sinx-1)
2. Let T be the Tension of the string
Weight of the mass=mg=3*9.8=29.4 N
Horizontal force acting on the mass by string=Tsinx=9.8 N
Vertical force by string=Tcosx=mg=29.4 N
Therefore Tsinx/Tcosx=tanx=9.8/29.4=1/3
cosx=3/√(3^2+1^2)=3/√10, sinx=1/√10
Weight of mass along the string
=mgcosx
=29.4*3/√10
=27.89 N
Weight of mass normal the string
=mgsinx
=29.4*1/√10
=9.30 N
To minimize the force, the external force should act perpendicular to T
The magnitude of the least force required is 9.30 N
3. Initially, P is hanging in equilibrium below O
-> P is initially 1m below O
When OP is horizontal, PQ=1-0.8=0.2m
To complete a circle about Q, the speed of P must be fast enough given the least centripetal force at the top being mg.
Therefore mg<=mv^2/r, v^2=0.2g
KE at the top
=0.5mu^2-mgh>=0.5mv^2
u^2>=0.2g+2g*(1+0.2)
u>=√(2.6g)
Minimum value of u = √(2.6g)
2013-06-03 15:48:53 補充：
In order not to exceed the upper character limit, I have to clump many expressions by eliminating spaces or unnecessary characters. Sorry for the inconvenience.
收錄日期: 2021-04-16 15:56:07
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https://hk.answers.yahoo.com/question/index?qid=20130530000051KK00177