physics

A 3.0kg- block sits on top of a 5.0kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force as shown in the figure . The coefficient of static friction between all surfaces is 0.65 and the kinetic coefficient is 0.42.

Part a: What is the minimum value of needed to move the two blocks?

part b:

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  1. part b: If the force is 10% greater than your answer for (a), what is the acceleration of each block?

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  2. The forces acting on the blocks are:
    top block (m1) :
    Gravity m1•g (downwards), the normal force on contact between two blocks N1 (upwards), friction force between two blocks F1(fr) ( to the right), Tension T (to the left).
    Bottom block (m2).
    Gravity m2•g (downwards), gravity m1•g (downwards), the normal force on contact between blocks and the surface N2 (upwards), friction force between two blocks F1(fr) (to the left), friction force between the block and the surface F2(fr) (to the left), tension T (to the left) , unknown force F (to the right).

    Equations of the blocks motion (projections on the horizontal and vertical axes):
    top block -
    x: m1•a = T – F1(fr),
    y: 0 = – m1•g + N1. => N1 = m1•g
    bottom block –
    x: m2•a = F –T –F1(fr) – F2(fr),
    y: 0 = - m1•g – m2•g +N2. => N2 = (m1+m2) •g

    (a) If the two blocks are just to move, then the force of static friction will be at its maximum, and sothe frictions forces are as follows.
    F1(fr) = k1• N1 = k1•m1•g,
    F2(fr) = k1• N2 = k1•(m1+m2)•g,

    a = 0;
    0 = T – F1(fr), => T = F1(fr),
    0 = F –T –F1(fr) – F2(fr),
    Solve for F.
    F = T +F1(fr) + F2(fr) = F1(fr) +F1(fr) + F2(fr) = 2 •F1(fr) + F2(fr) =
    = 2• k1•m1•g + k1•(m1+m2)•g = k1•g (3•m1+m2) = 0.65•9.8•(3•3+5) = 89.18 N.

    (b) Now force is F1 = 1.1•89.18 = 98.1 N
    m1•a = T – F1(fr),
    m2•a = F1 –T –F1(fr) – F2(fr),
    (m1+m2) •a = F1 - 2 •F1(fr) - F2(fr) =F1 –k2•g•(3•m1+m2),

    a = {F1 –k2•g•(3•m1+m2)}/(m1+m2) =
    ={98.1–0.42•9.8•(3•3+5)}/(3+5) =
    = 5.06 m/s².

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