Two blocks of mass m 1 = 12 kg and m 2 = 8 kg are sliding down an inclined plane

Question

Two blocks of mass m1 = 12 kg and m2 = 8 kg are sliding down an inclined plane (see figure below).

(a) If the plane is frictionless, what is the magnitude of the contact force between the two masses? 0 Correct: Your answer is correct. N

(b) If the coefficients of kinetic friction between m1 and the plane is ?1 = 0.25 and between m2 and the plane is ?2 = 0.25, what is the contact force between the two masses? 0 Correct: Your answer is correct. N

(c) If ?1 = 0.16 and ?2 = 0.25, what is the contact force? 1.714 Incorrect: Your answer is incorrect. N

Explanation / Answer

for m1

m1*g*sintheta - f = m1*a........(1)

for m2

m2*g*sintheta + f = m2*a .......(2)

1 + 2

(m1+m2)*g*sintheta = (m1+m2)*a

a = g*sintheta

from f = m1*g*sintheta - m1*a

f = m1*g*sintheta - m1*g*sintheta = 0

------------------

(b)

for m1

m1*g*sintheta - f - uk1*m1*g*costheta = m1*a.......(1)

for m2

m2*g*sintheta + f - uk2*m2*g*costheta = m2*a............(2)

1 + 2

a = g*sintheta - (g*costheta*(uk1*m1 + uk2*m2)/(m1+m2))

from 1

f = m1*g*sintheta - uk1*m1*g*costheta - m1*a

f = m1*g*sintheta - uk1*m1*g*costheta - m1*g*sintheta + (uk1*m1^2*g*costheta + uk2*m1*m2*g*costheta)/(m1+m2)

f = -uk1*m1*g*costheta + m1*g*costheta*(uk1*m1 + uk2*m2)/(m1+m2)

f = (-uk1*m1^2*g*costheta - uk1*m1*m2*g*costheta + uk1*m1^2*g*costheta + uk2*m1*m2*g*costheta )/(m1+m2)

f = m1*m2*g*costheta*(uk2-uk1)/(m1+m2)

uk1 = uk2 = 0.25

contact force f = 0

------------------------

(c)

contact force f = 12*8*9.8*cos35*(0.25-0.16)/(12+8) = 3.468 N

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