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!