A mass M kg is suspended by a weightless
string. The horizontal force required to hold
the mass at 60° with the vertical is
1) Mg
2) Mg root 3
3) Mg(root3+1)
4)Mg / root 3
Pls tell the correct answer
string. The horizontal force required to hold
the mass at 60° with the vertical is
1) Mg
2) Mg root 3
3) Mg(root3+1)
4)Mg / root 3
Pls tell the correct answer
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Work done by tension + Work done by force (applied) + Work done by gravitational force = change in kinetic energy
Work done by tension is zero
⇒0+F×AB−Mg×AC=0
⇒F=Mg(ACAB)=Mg⎡⎣⎢1−12√12√⎤⎦⎥
[∵AB=lsin45∘=l2–√
and AC=OC−OA=l−lcos45∘=l(1−12–√)
where l= length of the string.]
F⇒Mg(2–√−1)
Answer:
answer is option 2) Mg root 3