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 2 of 2.
Srikanth said:
1 decade ago
Can you please explain how does the tension in AB and BC became same i.e. T ?
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
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
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.
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.
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