Laws of Motion 3 Question 5

5. A uniform wooden stick of mass 1.6kg of length l rests in an inclined manner on a smooth, vertical wall of height h(<l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30 with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio hl and the frictional force f at the bottom of the stick are (g=10ms2)

(a) hl=316,f=1633N

(b) hl=316,f=1633N

(c) hl=3316,f=833N

(d) hl=3316,f=1633N

(2016 Adv.)

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Answer:

Correct Answer: 5. (d)

Solution:

N1cos30f=0

N1sin30+N2mg=0

mgl2cos60N1hcos30=0

Also, given N1=N2

Solving Eqs. (i), (ii), (iii) and (iv) we have

hl=3316 and f=1633



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