Three Boxes are located on three inclined surfaces with friction. The friction c

Question

Three Boxes are located on three inclined surfaces with friction. The friction coefficients are identical. Angles of inclined surfaces with the horizontal direction Theta_1, Theta_2, and Theta_3 are different. On the inclined surface with the angle Theta_1, the acceleration of the box is a_1. On the inclined surface with the angle Theta_2, the acceleration is a_2. The motion on the third surface happens without any acceleration. Find an expression for third angle in terms of Theta1, a_1, Theta_2, a_2.

Explanation / Answer

Here, For first incline

=>    ma1 = mgsin(1) - umgcos(1)

For second incline

=>    ma2 = mgsin(2) - umgcos(2)

For third incline

=>    mgsin(3) = umgcos(3)

=>   tan(3) = u

=>   ma1 = mgsin(1) - tan(3) * mgcos(1)

=>    a1 =   gsin(1) - tan(3) * gcos(1)

=>     3   = tan-1 [gsin(1) - a1]/(gcos(1))

Also,    ma2 = mgsin(2) - tan(3) * mgcos(2)

=>     a2 = gsin(2) - tan(3) * gcos(2)

=>     3   = tan-1 [gsin(2) - a2]/(gcos(2))

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