# 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.

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1. 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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