Question

Determine the horizontal force required to displace a mass of 0.03kg suspended by a string until the string makes an angle 30* with the vertical?

Answer 1

Given,
m = 0.03 kg
Using the law of conservation of mechanical energy,
Work done by the horizontal force = Gain in P. E. of mass m
or, F × 1 sin 30° = mg × ( 1 - 1 cos 30°)
or, F = mg × (1 -1 cos 30°)/1 sin 30°
= 0.03 g × (1 - √3/2)/(1/2)
= 0.03 g × (2 - √3/2)/(1/2)
= 0.03 g × (2 - √3)
= 0.03 g × 9.8 × (2 - √3)
= 0.079 N Answer

Answer 2

see diagram.

Let the tension force in the string be T.  Let the force required be F.   The static equilibrium of pendulum (mass) at an angle Ф requires:

 F = T SinФ                     W =  m g = T CosФ       =>  F = m g Tan Ф

As Ф increases, TanФ increases and so Force also increases.  So the force required maximum at Ф = 30°.
        F = 0.03 kg * Tan 30° * g   = √3/100  kg wt    or √3 g /100  Newtons
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If we want the average force along the horizontal direction (x) to move the mass to 30 deg., then we find that by integration.

$Avg.\ Force=\frac{1}{x} \int \limits_{0}^{\phi}\ {F} \, dx\\\\x=L\ Sin\phi,\ dx=LCos\phi,\ \ F=mg\ Tan\phi\\\\Avg.\ Force=\frac{1}{LSin\phi} \int \limits_{0}^{\phi}\ {mgTan\phi*cos\phi} \, d\phi\\\\=mg\frac{1-Cos\phi}{sin\phi},\ \ or,\ mgTan\frac{\phi}{2}\\\\=0.03*10*(2-\sqrt3)\ N,\ \ for\ \phi=30^o\\\\=0.08 N\ (avg)$