A uniform board of length L and mass M lies near a boundary that separates two r

Q: A uniform board of length L and mass M lies near a boundary that separates two r

? let position of the board on top picture x=0; now if x>0 then force of friction f(x)= ?1*m1*g + ?2*m2*g, where m2=(M/L)*x, m1= (M/L)(L-x);thus ? f(x)=(Mg/L) *(?1*(L-x) + ?2*x) = (Mg/L) *(?1*L +( ?2-?1)*x); ? elementary work dw= f*dx= (Mg/L) (?1*L +(?2-?1)x)dx, hence ? w=(Mg/L) *(?1*L*x +0.5(?2-?1)x^2) = = {for x=0 until L} =(Mg/L) *(?1*L^2 +0.5(?2-?1)L^2) = = 0.5MgL *(?1+?2);

ID: 2143587 • Letter: A

Question

A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is ?1, and in region 2, the coefficient is ?2. The positive direction is shown in the figure.

Explanation / Answer

? let position of the board on top picture x=0;
now if x>0 then force of friction f(x)= ?1*m1*g + ?2*m2*g, where
m2=(M/L)*x, m1= (M/L)(L-x);thus
? f(x)=(Mg/L) *(?1*(L-x) + ?2*x) = (Mg/L) *(?1*L +( ?2-?1)*x);
? elementary work dw= f*dx= (Mg/L) (?1*L +(?2-?1)x)dx, hence
? w=(Mg/L) *(?1*L*x +0.5(?2-?1)x^2) =
= {for x=0 until L} =(Mg/L) *(?1*L^2 +0.5(?2-?1)L^2) =
= 0.5MgL *(?1+?2);

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A uniform board of length L and mass M lies near a boundary that separates two r

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