Question #124626
A particle of mass 10 kg is attached to one end of a light inextensible string whose other end
is fixed. The particle is pulled aside by a horizontal force until the string is at 60° to the
vertical. Find the magnitude of the horizontal force and the tension in the string.
1
Expert's answer
2020-07-01T18:22:56-0400


According to second Newton's Law


F+mg+T=0\vec{F}+m\vec{g}+\vec{T}=0


Oy:Tcosθmg=0Oy: T\cos\theta-mg=0

Ox:FTsinθ=0Ox: F-T\sin\theta=0


T=mgcosθT={mg\over \cos \theta}

F=mgtanθF=mg\tan \theta

T=10kg9.81m/s2cos60°=196.2 NT={10kg\cdot9.81 m/s^2\over \cos 60 \degree}=196.2\ N

F=10kg9.81m/s2tan60°=98.13 N169.9 NF=10kg\cdot9.81m/s^2\tan60\degree=98.1\sqrt{3} \ N\approx169.9\ N



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