0
1.5kviews
A heavy metal bar AB rests with its lower end A on a rough horizontal floor having coefficient of friction μF & the other end b on a rough vertical wall having coefficient of friction μw.

If the the centre of gravity of the bar is at distance a and b from the ends A & b respectively, show that at impending motion, the inclination of thee bar with the horizontal will be:

θ=tan1(1μFabμFμWa+b)

enter image description here

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2016

1 Answer
1
18views

fx=0

N2μFN1=0N2=μFN1(1)fy=0μwN2+N1W=0

From (1)

μw.μFN1+N1=WN1=W1+μwμF and N2=μF1+μwμFWMA=0w×acosθμwN2(a+b)cosθN2(a+b)sinθ=0cosθ.[W.a(a+b)μw.μF1+μwμFW]=μf1+μwμFW(a+b)sinθsinθcosθ=[aμwμF(a+b)1+μwμfμF1+μwμF(a+b)]θ=tan1[a(1+μFμw)(a+b)μwμFμF(a+b)]θ=tan1[a+aμFμwaμFμwbμFμwμF(a+b)]θ=tan1[1μF.abμFμw(a+b)]

Hence proved.

Please log in to add an answer.