Answer to Question #164738 in Mechanics | Relativity for Christian Curry

Question

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

Answer 1

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

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 99.7N

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 64.9N

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 60N

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 149.6N

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 97.4N

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

$(m_1g)^2 = 2T^2 - 2T^2cos\theta_1 = 2T^2(1-cos\theta_1) \to T = \large\frac{m_1g}{\sqrt{2(1-cos\theta_1)}}$ = 90N