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Answer to Question #20007 in Mechanics | Relativity 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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#340153 on Dec 2023
#340153 on Dec 2023
Comments
Thank you for attention! We replaced the wrong answer.
THIS 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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