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.
Kaliraj said:
5 years ago
I got the θ value as 18.34°.
How all getting 23°? Please explain to me.
How all getting 23°? Please explain to me.
Siva said:
6 years ago
How do you find bc = 1.25?
Balaji said:
7 years ago
(angle)BCD= 23.578.
Then calculating the tension in string BC. Let it be called T.
Now T is affected totally by the 10lb block.
So,
T= 10cos= 10cos23.578=9.1651.
(Note : We're taking 10cos' because only the cos component of the force due to 10lb block is acting in string BC).
Tension in AB = Tension in BC =T.
Mass of E = T in AB +T in BC,
So, 9.1651+9.1651= 18.3302 ~ 18.33lb.
Then calculating the tension in string BC. Let it be called T.
Now T is affected totally by the 10lb block.
So,
T= 10cos= 10cos23.578=9.1651.
(Note : We're taking 10cos' because only the cos component of the force due to 10lb block is acting in string BC).
Tension in AB = Tension in BC =T.
Mass of E = T in AB +T in BC,
So, 9.1651+9.1651= 18.3302 ~ 18.33lb.
Ram said:
8 years ago
Anyone, can explain clearly?
Prasanna said:
10 years ago
I didn't understand you are explanation.
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.
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.
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
Nagu said:
1 decade ago
Why we have to double the value ?
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
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