IIT JEE , AIEEE Forum - Physics , Mechanics

nishant_88

Please tell me the answer of q. No 22 Pg. 115 of HC Verma Part 1.

Show the working also.

metal

The solutions:-

a.)At B:-

normal contact force + mv²/r = mg

or, normal contact force=n = m(g- v²/r)=975 N

At D:-

n=mg+mv²/r=1025 N

b.)friction at B and D are = 0 N [no acceleration]

friction at C=mgcos#, where # is the angle made with the tangent drawn at C with th vertical and # = 45⁰

so,friction at C=707 N [approx.]

c.)Just before C:-

n+mv²/r = mgcos#

n=682 N

Just after C:-

n=mv²/r+mgcos#

n=732 N

d.)minimum frictional coefficient = 707/682= 1.037 [approx.]

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netkid07

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