Answer:1135.13 N-m
Step-by-step explanation:
Given
inclination of tree from horizontal [tex]\theta =27^{\circ}[/tex]
mass of first child [tex]m_1=40 kg[/tex]
distance of first child [tex]r_1=1.8 m[/tex]
mass of first child [tex]m_2=29 kg[/tex]
distance of first child [tex]r_2=2.0 m[/tex]
Torque exerted by first child
[tex]T_1=m_1g\cos \theta \times r_1[/tex]
[tex]T_1=40\times 9.8\cos 27\times 1.8=628.69 N-m[/tex]
Torque exerted by second child
[tex]T_2=m_2g\cos \theta \times r_2[/tex]
[tex]T_2=29\times 9.8\cos 27\times 2=506.44 N-m[/tex]
Both torques are in same sense therefore
[tex]T_{net}=T_1+T_2[/tex]
[tex]T_{net}=628.69+506.44[/tex]
[tex]T_{net}=1135.13 N-m[/tex]
