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.
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.
To calculate the value of the applied force in each scenario, we need to consider the forces acting on the crates and apply Newton's second law of motion.
(a) (i) When the crates travel at constant velocity, the applied force is equal to the force of friction experienced by the crates. The force of friction can be calculated using the formula:
F_friction = μ * N
where μ is the coefficient of friction and N is the normal force acting on the crates. The normal force can be calculated as the weight of the crates, which is equal to the mass multiplied by the acceleration due to gravity (g = 9.8 m/s²).
For Crate A:
F_friction(A) = μ(A) * N(A)
= μ(A) * m(A) * g
For Crate B:
F_friction(B) = μ(B) * N(B)
= μ(B) * m(B) * g
The total force of friction acting on both crates is the sum of the individual forces of friction:
F_friction(total) = F_friction(A) + F_friction(B)
Since the crates are traveling at constant velocity, the applied force is equal to the total force of friction:
F_applied = F_friction(total)
(ii) When the crates accelerate to the right at 0.3 m/s², the applied force needs to overcome the force of friction and also provide the necessary force to accelerate the crates. The force of friction remains the same as in part (i).
The net force required to accelerate the crates can be calculated using the formula:
Net force = (m(A) + m(B)) * a
where a is the acceleration of the crates. Substitute the given values to find the net force.
The applied force must be equal to the net force:
F_applied = Net force
(b) If the crates accelerate at 0.15 m/s² to the right and the force is applied to Crate B at an angle of 25° anticlockwise to the horizontal, we need to consider the horizontal and vertical components of the applied force.
The horizontal component of the applied force, F_applied_horizontal, is equal to:
F_applied_horizontal = F_applied * cos(25°)
The vertical component of the applied force, F_applied_vertical, is equal to:
F_applied_vertical = F_applied * sin(25°)
The horizontal component of the applied force contributes to the acceleration of the crates, while the vertical component only affects the normal force.
Using the formula for net force in the horizontal direction:
Net force_horizontal = (m(A) + m(B)) * a_horizontal
where a_horizontal is the horizontal component of the acceleration of the crates. Substitute the given values to find the net force_horizontal.
The net force_horizontal must be equal to the horizontal component of the applied force:
Net force_horizontal = F_applied_horizontal
Thus, you can now solve for the value of the applied force.