Question Number 102978 by ajfour last updated on 12/Jul/20

Commented byajfour last updated on 12/Jul/20

The inclined plane of mass m,  length L and the block of mass m,  both are on the verge of slipping.  Find x and μ.

Answered by mr W last updated on 12/Jul/20

Commented bymr W last updated on 12/Jul/20

x=L sin θ  C=center of total mass of both objects  M=2m  BC=p=((mx+m×(L/2))/(2m))=((L(2 sin θ+1))/4)  CA=q=L−p=((L(3−2 sin θ))/4)  tan φ=μ  such that the block m doesn′t slip,  tan θ=μ  ⇒φ=θ  α=(π/2)−θ−φ=(π/2)−2θ  β=φ+θ=2θ  ((SC)/(sin β))=((BC)/(sin ((π/2)−φ)))=(p/(cos θ))  ⇒SC=((p sin β)/(cos θ))=((L(2 sin θ+1) sin 2θ)/(4 cos θ))  =((L(2 sin θ+1) sin θ)/2)  ((SC)/(sin α))=((CA)/(sin φ))=(q/(sin θ))  ⇒SC=((q sin α)/(sin θ))=((L(3−2 sin θ) cos 2θ)/(4 sin θ))    ((L(3−2 sin θ) cos 2θ)/(4 sin θ))=((L(2 sin θ+1) sin θ)/2)  ⇒8 sin^2  θ+2 sin θ−3=0  ⇒(2 sin θ−1)(4 sin θ+3)=0  ⇒sin θ=(x/L)=(1/2) or −(3/4) (rejected)  ⇒θ=30°  ⇒μ=tan θ=(1/(√3))

Commented bymr W last updated on 12/Jul/20

if block and plane have different  masses, say mass of block is λm, and  mass of plane is m, then  M=(1+λ)m  BC=p=((λmx+m×(L/2))/((λ+1)m))=((L(2λ sin θ+1))/(2(λ+1)))  CA=q=L−p=((L[1+2λ(1−sin θ)])/(2(λ+1)))  ⇒SC=((p sin β)/(cos θ))=((L(2λ sin θ+1) sin θ)/((λ+1)))  ⇒SC=((q sin α)/(sin θ))=((L[1+2λ(1−sin θ)]cos 2θ)/(2(λ+1) sin θ))  ((L[1+2λ(1−sin θ)]cos 2θ)/(2(λ+1) sin θ))=((L(2λ sin θ+1) sin θ)/((λ+1)))  [1+2λ(1−sin θ)](1−2 sin^2  θ)=2(2λ sin θ+1) sin^2   with t=sin θ  4(λ+1)t^2 +2λt−(2λ+1)=0  t=((−2λ+(√(4λ^2 +16(λ+1)(2λ+1))))/(8(λ+1)))  t=((−λ+3λ+2)/(4(λ+1)))=(1/2)=sin θ  that means the solution is the same  even when both objects have different  masses.

Commented byajfour last updated on 12/Jul/20

Great!  I got chance to understand  the principle once again Sir.  Thanksfor this viewpoint.

Commented byajfour last updated on 12/Jul/20

Nice, what made you suspect so  Sir ?

Commented bymr W last updated on 12/Jul/20

for θ=30° we can see that both  masses are at the same point, that  means it doesn′t matter how large  the masses are, i.e. θ=30° is always  a solution for any masses. i wanted   know if other solution is possible.  now i know it is not possible.

Answered by ajfour last updated on 12/Jul/20

Commented byajfour last updated on 12/Jul/20

F=mgcos θ  And as block is on verge of slipping     f=μF = μmgcos θ=mgsin θ  ⇒   μ=tan θ       .....(i)  ⇒  sin θ=(μ/(√(1+μ^2 )))      ....(ii)  system: block+inclined plane_(−)      N=μR         (ΣF_x =0)     .....(iii)      μN+R=2mg     (ΣF_y =0)  ⇒    μ^2 R+R=2mg  ⇒    ((mg)/R)=((μ^2 +1)/2)       .....(iv)   Torque about position of block_(−)     (μNcos θ+Nsin θ)x  −(mgcos θ)(x−(L/2))       = (Rcos θ−μRsin θ)(L−x)  Now dividing by RLcos θ  and using eqs. (i)→(iv) namely    (N/R)=μ  ,  ((mg)/R)=((μ^2 +1)/2) , tan θ=μ ;     (x/L)=sin θ=(μ/(√(μ^2 +1)))   we arrive at    μ(μ+μ)((μ/(√(μ^2 +1))))    −(((μ^2 +1)/2))((μ/(√(μ^2 +1)))−(1/2))   = (1−μ^2 )(1−(μ/(√(μ^2 +1))))  or in terms of θ  2tan ^2 θsin θ−(((sin θ−(1/2)))/(2cos ^2 θ))        =(1−tan ^2 θ)(1−sin θ)  let  sin θ=t   ⇒  2t^3 −(((2t−1))/4)=(1−2t^2 )(1−t)  ⇒  2t^3 −(t/2)+(1/4)=1+2t^3 −t−2t^2   ⇒   8t^2 +2t−3=0  ⇒   8(t+(1/8))^2 =((25)/8)  ⇒  t=−(1/8)±(5/8)  ⇒t= sin θ=(1/2)   &  μ=tan θ=(1/(√3)) ■

Commented bymr W last updated on 12/Jul/20

wonderful solution sir!