Question #164738

Find the tensions in the ropes for each case. Note that theta 1= 35, theta 2= 55, theta 3=60, m1= 6 kg, and m2=9 kg


1
Expert's answer
2021-02-19T10:30:28-0500

The picture must be given, so I imagine the picture.



1) θ1=35o m1=6kg\theta_1 = 35^o \space m_1 = 6kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 99.7N

2) θ1=55o m1=6kg\theta_1 = 55^o \space m_1 = 6kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 64.9N

3) θ1=60o m1=6kg\theta_1 = 60^o \space m_1 = 6kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 60N

4) θ1=35o m1=9kg\theta_1 = 35^o \space m_1 = 9kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 149.6N

5) θ1=55o m1=9kg\theta_1 = 55^o \space m_1 = 9kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 97.4N

6) θ1=60o m1=9kg\theta_1 = 60^o \space m_1 = 9kg

(m1g)2=2T22T2cosθ1=2T2(1cosθ1)T=m1g2(1cosθ1)(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}} = 90N


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