physics
posted by Nick .
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. xaxis is directed to the right, yaxis is directed upwards.
Part a.
Freebody 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 xaxis are:
0 = T  F1(fr),.........(1)
0 =  T  F2(fr) + F,...(2)
Projections on yaxis 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 yaxis 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 yaxis 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.
Respond to this Question
Similar Questions

Physics
Sally applies a horizontal force of 500 N with a rope to dtag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 67.5 degrees. a) How much force is exerted by the rope on the crate? 
physics
Two wooden crates with masses are as shown are tied together by a horizontal cord. Another cord is tied to the first crate and it is pulled with a force of 199N at a angle of 20.0 degrees. Each crate has a coefficient of kinetic friction … 
Physics
A crate with a total mass of 18.0 kg is pulled by a person along the floor at a constant speed by a rope. The rope is inclined at 20 degree above the horizontal, and the crate moves 20.0 m on a horizontal surface. The coefficient of … 
Physics
A crate of mass 100kg is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.3. A horizontal force, F, or magnitude 350N is applied … 
Physics Force and Friction question
A crate of mass 100kg is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.3. A horizontal force, F, or magnitude 350N is applied … 
university physics
there is a crate of mass 4.67 kg on top of another crate of mass 2.29 kg. The coefficient of friction between the lower crate and the floor is μk = 0.280 and the coefficient of static friction between the two crates is μs … 
physics
Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction … 
physics
Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction … 
physics
Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction … 
physics
Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction …