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12 Sep 2007 01:20:20 IST

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friction question plz solve with proper explanation

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The handle of a floor mop of mass m makes an angle ? with the vertical direction. Let µk be the coefficient of kinetic friction between mop and floor and µs be the coefficient of static friction between mop and floor. Neglect the mass of the handle. (a) Find the magnitude of the force F direction along the handle required to slide the mop with uniform velocity across the floor. (b) Show that if ? is smaller than a certain angle ??, the mop cannot be made to slide across the floor no matter how great a force is directed along the handle. (c) What is the angle ???

Answer

(a) F= (µ_k mg)/ (sin?-µ_kcos?)

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Comments (5)

Amrita bhavaraju

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Joined: 28 May 2007

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12 Sep 2007 14:11:29 IST

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whats the ans.???????????

Amrita bhavaraju

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Joined: 28 May 2007

Posts: 107

13 Sep 2007 11:52:56 IST

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heyy....i got it....

Amrita bhavaraju

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Joined: 28 May 2007

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13 Sep 2007 12:05:44 IST

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here fcos

b also acts along with mg as normal reaction so while it is moving, the frictional force is

k(fcos

+mg)...as it is moving with uniform velocity fsin

k(fcos

+mg)... simplifying we get f=(

kmg/sin

kcos

)......

Amrita bhavaraju

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Joined: 28 May 2007

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13 Sep 2007 12:44:00 IST

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for the mop to have motion fsin

s(fcos

+mg)...for angle less than tan-1

, sin

scos

<0 ==>f is negative ..so for an angle less than

=tan^{-1 _s,} the mop will not move however large the force may be.............if u find it useful plzzzzzzzzzzz rate ..:)

raymont cherian

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Joined: 3 Jul 2007

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23 Sep 2007 21:25:12 IST

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the numerator of the ans is correct . bt the denominator is wrong.. it wud be cos?-

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