Answer to Question #124504 in Mechanics | Relativity for Emmanuel Kojo Mensah Kent

Question #124504

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
2020-07-02T17:52:30-0400

The sum of forces' vectors is 0\overrightarrow{0} : R=ma=F+T+mg=0\overrightarrow{R} = m\overrightarrow{a} = \overrightarrow{F} + \overrightarrow{T} + m\overrightarrow{g} = \overrightarrow{0}

Horizontal direction: F=Tsin60;F = T \cdot sin60^\circ;

Vertical direction: Tcos60=mg;T \cdot cos 60^\circ = mg;

Then T=mgcos60=mg1/2=2mg;T = \dfrac{mg}{cos60^\circ} = \dfrac{mg}{1/2} = 2mg; F=Tsin60=2mg32=3mgF = T \cdot sin60^\circ = 2mg \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}mg

m is known from the task and g you should take with the necessary accuracy.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment