Engineering Mechanics - Equilibrium of a Particle - Discussion
Discussion Forum : Equilibrium of a Particle - General Questions (Q.No. 2)
2.
A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft.
Discussion:
15 comments Page 1 of 2.
Pratik/Satish said:
1 decade ago
Considering traingle BCD between BC and CD is 23.57 degrees.so 10cos(23.57)= 9.165, double of which is the value of weight of block E i.e. 18.33.
Ajinkya said:
1 decade ago
Hey guys can you explain how did you get angle between bc and cd as 23.57 degrees ? please reply.
Sidhu said:
1 decade ago
Half of the distance of AC is 0.5 feet = BD
and CD = 1.5 ft
and CD = 1.5 ft
MMWaris said:
1 decade ago
Sidhu, let me explain you.
its BC= 1.25ft (={4-1.5}/2), which will be used to calculate sin(theta) together with BD=0.5ft.
i.e. Sin(theta) = 0.5/1.25=0.4
From here cos(theta) will be calculated. (theta is the angle at C )
Cos(theta)=sqrt{1-(sin(theta))square}=0.916
Then Tension in the rope
T=M(D)*g -(1)
And at E, the forces are
M(E)*g=2*T*Cos(Theta) -(2)
from (1) & (2)
M(E)=2*M(D)*Cos(theta)=2*10*0.916=18.32 lb
its BC= 1.25ft (={4-1.5}/2), which will be used to calculate sin(theta) together with BD=0.5ft.
i.e. Sin(theta) = 0.5/1.25=0.4
From here cos(theta) will be calculated. (theta is the angle at C )
Cos(theta)=sqrt{1-(sin(theta))square}=0.916
Then Tension in the rope
T=M(D)*g -(1)
And at E, the forces are
M(E)*g=2*T*Cos(Theta) -(2)
from (1) & (2)
M(E)=2*M(D)*Cos(theta)=2*10*0.916=18.32 lb
Srikanth said:
1 decade ago
Can you please explain how does the tension in AB and BC became same i.e. T ?
Raunak said:
1 decade ago
Rope is of 4ft
& ab=bc.... bc=(4-1.5)x1/2
=1.25
& by hypo... bd=0.5
now let angle at c=θ
sin(θ)=0.5/1.25=0.4
cos(θ)=sqrt(1-sin2θ)=0.916
now, T,TENSION in rope
T=MxDxg-------------(1)
Forces at E,
MxExg=2Tcosθ--------(2)
from 1 & 2 we get
MxE=2xMxDxcosθ
=2x10x0.916
=18.32
& ab=bc.... bc=(4-1.5)x1/2
=1.25
& by hypo... bd=0.5
now let angle at c=θ
sin(θ)=0.5/1.25=0.4
cos(θ)=sqrt(1-sin2θ)=0.916
now, T,TENSION in rope
T=MxDxg-------------(1)
Forces at E,
MxExg=2Tcosθ--------(2)
from 1 & 2 we get
MxE=2xMxDxcosθ
=2x10x0.916
=18.32
Nagu said:
1 decade ago
Why we have to double the value ?
Yuel said:
1 decade ago
You can determine the weight of the suspended block 'E' BY USING COSINE LAW IN FORCE TRIANGLE.
applying the law of cos in the triangle, we will obtain the ff:
c = E
a = 10
b = 10
∠= 132.84
where ∠= 2θ - 180
∠= 2(23.58) - 180
∠= 132.84
substitute the value in cosine law where
c^2 = a^2 + b^2 - 2abcosâˆ
E^2 = 10^2 + 10^2 - 2(10)(10)cos132.84
E = 18.33
XD
applying the law of cos in the triangle, we will obtain the ff:
c = E
a = 10
b = 10
∠= 132.84
where ∠= 2θ - 180
∠= 2(23.58) - 180
∠= 132.84
substitute the value in cosine law where
c^2 = a^2 + b^2 - 2abcosâˆ
E^2 = 10^2 + 10^2 - 2(10)(10)cos132.84
E = 18.33
XD
Bala said:
1 decade ago
@Sri kanth, I can explained them. AB=BC THAT STRUCTURE is triangle. Two side ninety degree in ago. A to C DISTANCE less than. Touch the the both way of two side return the value same.
AJEET SONI,ABBAGONI MAHESH said:
10 years ago
Cos A = 0.5/1.25.
Then a = 66.42.
According to Lamis theorem We/sin (23.58+23.58) = 10/sin (90+66.42).
We = 10xsin 47/sin 156 = 18.33.
Then a = 66.42.
According to Lamis theorem We/sin (23.58+23.58) = 10/sin (90+66.42).
We = 10xsin 47/sin 156 = 18.33.
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