Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A stick of mass density λ=8 kg/m rests on a disc of radius R = 20cm as shown in the figure. The stick makes an angle θ=37o with the horizontal and is tangent to the disc at its upper end. Friction exists at all points of contact and assume that it is large enough to keep the system at rest. Find the friction force (in Newton) between the ground and the disc. (take g  = 10  m/s2 )

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 6.40.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Since, disc is in rotational equilibrium,friction due to rod and ground will be same.τA=0At point of contact of ground⇒NL=MgL2cos⁡θ⇒N=Mg2cos⁡θ For disc, Nsin⁡θ=f+fcos⁡θ∴f=Nsin⁡θ1+cos⁡θ=Mgsin⁡θcos⁡θ2(1+cos⁡θ)M=λL and L=Rtan⁡θ2⇒f=12λRgcos⁡θ =12×8×10×210×45=325=6.40N
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon