Thursday

October 27, 2016
Posted by **Nick** on Sunday, June 3, 2012 at 6:56pm.

(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 -
**Elena**, Monday, June 4, 2012 at 5:37amMake 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.