### Discussion :: Equilibrium of a Particle - General Questions (Q.No.2)

Pratik/Satish said: (Aug 28, 2011) | |

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: (Sep 28, 2011) | |

Hey guys can you explain how did you get angle between bc and cd as 23.57 degrees ? please reply. |

Sidhu said: (Oct 4, 2011) | |

Half of the distance of AC is 0.5 feet = BD and CD = 1.5 ft |

Mmwaris said: (Oct 6, 2011) | |

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 |

Srikanth said: (Oct 20, 2011) | |

Can you please explain how does the tension in AB and BC became same i.e. T ? |

Raunak said: (Nov 19, 2011) | |

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 |

Nagu said: (Jun 14, 2012) | |

Why we have to double the value ? |

Yuel said: (Aug 22, 2012) | |

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 - 2abcos∝ E^2 = 10^2 + 10^2 - 2(10)(10)cos132.84 E = 18.33 XD |

Bala said: (Jul 22, 2013) | |

@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: (Sep 29, 2015) | |

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. |

Prasanna said: (Oct 30, 2015) | |

I didn't understand you are explanation. |

Ram said: (May 25, 2017) | |

Anyone, can explain clearly? |

Balaji said: (Jun 9, 2018) | |

(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. |

Siva said: (Aug 17, 2019) | |

How do you find bc = 1.25? |

Kaliraj said: (Aug 12, 2020) | |

I got the θ value as 18.34°. How all getting 23°? Please explain to me. |

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