# Answer on Mechanics | Relativity Question for Koid Xian Ting

Question #20007

An ant wants to take a big piece of candy with mass m=5 g back to its colony. Find the minimum force it needs to move the candy piece (in mN).

Details and assumptions

The friction coefficient between the candy and the ground is k=0.5

The gravitational acceleration is g=9.8 m/s2^2

Hi! The above question disappointed me. Because it does not state whether it is static or kinetic friction coeffiecient. I gave 2 answers 24.5 and 24.6 but i get wrong. Please help me. Thanks

Details and assumptions

The friction coefficient between the candy and the ground is k=0.5

The gravitational acceleration is g=9.8 m/s2^2

Hi! The above question disappointed me. Because it does not state whether it is static or kinetic friction coeffiecient. I gave 2 answers 24.5 and 24.6 but i get wrong. Please help me. Thanks

Expert's answer

For minimum force,

F cos a = f [eqn 1]

where F is the force required, a is the optimum angle for minimum, force f is the friction force.

f is related to F and mg as

f = ( mg - F sin a ) k [eqn 2 ]

where k represents the coefficient of friction. Equiting the two equations

F cos a = ( mg - F sin a ) k

Solving for F,

F = (mgk)/( ( cos a ) + k sin a ) [eqn 3]

For min.force , dF/da = 0 Differentiating equation 3 with respect to a, dF/da = -mgk*([k cos a ] - sin a)/(cos a + k sin a )^2

Solving for a when dF/da equal to 0 , a = 26.5650512° Substituting a into eqn 3 , F = 0.0219 N or 21.9 mN

F cos a = f [eqn 1]

where F is the force required, a is the optimum angle for minimum, force f is the friction force.

f is related to F and mg as

f = ( mg - F sin a ) k [eqn 2 ]

where k represents the coefficient of friction. Equiting the two equations

F cos a = ( mg - F sin a ) k

Solving for F,

F = (mgk)/( ( cos a ) + k sin a ) [eqn 3]

For min.force , dF/da = 0 Differentiating equation 3 with respect to a, dF/da = -mgk*([k cos a ] - sin a)/(cos a + k sin a )^2

Solving for a when dF/da equal to 0 , a = 26.5650512° Substituting a into eqn 3 , F = 0.0219 N or 21.9 mN

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## Comments

Assignment Expert28.02.2014 09:10Thank you for attention!

We replaced the wrong answer.

SOYUZ26.02.2014 15:21THIS EXPERT DOESNOT KNOW ANYTHING...................................For minimum force , F cos a = f [eqn 1] Where F is the force required , a is the optimum angle for min.force f is the friction force

f is related to F and mg by : f = ( mg - F sin a ) k [eqn 2 ] Where k represents the coefficient of friction

Equiting the two equations F cos a = ( mg - F sin a ) k

Solving for F , F = (mgk)/( ( cos a ) + k sin a ) [eqn 3]

For min.force , dF/da = 0 Differentiating equation 3 with respect to a, dF/da = -mgk*([k cos a ] - sin a)/(cos a + k sin a )^2

Solving for a when dF/da equal to 0 , a = 26.5650512° Substituting a into eqn 3 , F = 0.0219 N or 21.9 mN

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