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Two crates A and B connected together by a light rope are pulled along a factory floor by an applied force, F. Crate A has a mass of 50 kg and the coefficient of friction (μ) between crate A and the floor is 0,25. Crate B has a mass of 35 kg and μ between B and the floor is 0,2.
(a) Calculate the value of the applied force if:
(i) the crates travel at constant velocity
(ii) if the crates accelerate to the right at 0,3 m/s2

(b) If the crates accelerate at 0,15 m/s2 to the right and F is now applied to crate B at an angle of 25° anticlockwise to the horizontal, calculate the value of the applied force.

  • physics -

    Make the drawing: from left to right: crate A , crate B. Force F is applied to the crate B horizontally to the right. x-axis is directed to the right, y-axis is directed upwards.

    Part a.
    Free-body diagram gives
    -the forces applied to the crate A: gravity m1•g (downwards),F1(fr) (to the left), normal force N1 (upwards), tension T (to the right).
    Since v=const, the veñtor sum of all forces is zero.
    -the forces applied to the crate B: gravity m2•g (downwards),F2(fr) (to the left), normal force N2 (upwards), tension T (to the left), F (to the right). Since v=const, the veñtor sum of all forces is zero.
    Projections of these forces on x-axis are:
    0 = T - F1(fr),.........(1)
    0 = - T - F2(fr) + F,...(2)
    Projections on y-axis are
    0 = - m1•g + N1, ==> N1= m1•g ...(3)
    0 =- m2•g + N2, ==> N2= m2•g ....(4).
    F1(fr) = μ1•N1 = μ1• m1•g ........(5)
    F2(fr) = μ2•N2 = μ2• m2•g .......(6)

    Add (1) and (2) and substitute (5) and (6)
    0 = - F1(fr) - F2(fr) + F.
    F = F1(fr) + F2(fr) =
    =g• (μ1• m1+μ2• m2) =
    =9.8(0.25•50 + 0.2•35) =19.11 N.

    Part b.

    The acceleration is the same for both crates, therefore,

    m1•a = T - F1(fr),...........(1)
    m2•a = - T - F2(fr) + F,.....(2)
    Projections on y-axis are
    0 = - m1•g + N1, ==> N1= m1•g ....(3)
    0 =- m2•g + N2, ==> N2= m2•g .....(4).
    F1(fr) = μ1•N1 = μ1• m1•g .........(5)
    F2(fr) = μ2•N2 = μ2• m2•g .........(6)

    Add (1) and (2) and substitute (5) and (6)
    a• (m1+m2) = - F1(fr) - F2(fr) + F.
    F = F1(fr) + F2(fr) + a• (m1+m2)
    =g• (μ1• m1+μ2• m2) + a• (m1+m2) =
    =9.8(0.25•50 + 0.2•35) +0.3(50+35) =
    =44.6 N.

    Part C
    m1•a = T - F1(fr),...............(1)
    m2•a = - T - F2(fr) + F•cosα,....(2)
    Projections on y-axis are
    0 = - m1•g + N1, ==> N1= m1•g ...(3)
    0 =- m2•g + N2 +F•sinα, ==>
    N2= m2•g - F•sinα ................(4).
    F1(fr) = μ1•N1 = μ1• m1•g ........(5)
    F2(fr) = μ2•N2 =
    =μ2• (m2•g - F•sinα ) ............(6)

    Add (1) and (2) and substitute (5) and (6)
    a• (m1+m2) = - F1(fr) - F2(fr) + F•cosα.
    F•cosα = μ1• m1•g + μ2• (m2•g - F•sinα ) + a• (m1+m2),
    F• (μ2• sinα + cosα) = g• (μ1• m1+μ2• m2) + a• (m1+m2),
    F={g• (μ1• m1+μ2• m2) + a• (m1+m2)}/ (μ2• sinα + cosα)=
    =0.15•85+9.8)0.25•50+0.2•35)=
    ={9.8(0.25•50+0.2•35)+0.3(50+35)}/
    (0.25•0.42+0.906)=
    = 24.9 N.

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