Question #146068

A small ball of mass O, 5 kg hangs on a non-stretchable cord. The cord is placed horizontally and then the ball is released. Find the tension force of the cord when it forms angle 30° with the vertical. The answer should be expressed in newtons [N]. Assume the free fall acceleration to be 10 m/s^2.


1
Expert's answer
2020-11-25T10:51:45-0500

Find the speed of the ball when the cord (length ll) makes a 30° angle:


v=2gh=2gl sin30°.v=\sqrt{2gh}=\sqrt{2gl\text{ sin}30°}.

Find the centripetal acceleration at this speed:


a=v2l=2g sin30°.a=\frac{v^2}{l}=2g\text{ sin}30°.

The acceleration creates a force of


F=ma=2mg sin30°.F=ma=2mg\text{ sin}30°.

The force of gravity that acts on the cord at this angle is


Fg=mg cos60°.F_g=mg\text{ cos}60°.

So, the tension is


T=F+Fg==mg[cos(90°30­°)+2sin30°]=7.5 N.T=F+F_g=\\=mg[\text{cos}(90°-30­°)+2\text{sin30°}]=7.5\text{ N}.


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