We again use the Principle of Superposition to solve this problem by separating the problem into forces acting along the horizontal or X-axis and those acting along the vertical or Y-axis. Then, for each axis, all forces will be along a line. We assign a positive or negative sign to the forces, depending on their direction along the axis, and then add them algebraically. There is no acceleration in the Y-direction, a_y = 0.  (F_net)_y = ma_y =m(0) = 0

(a) and (b),
F_N - F_g = 0  or  F_N = F_g = mg .
For m = 2.0 kg, F_N = F_g = (2.0 kg)(10 m/s²) = 20 N.

(F_net)_x = ma_x
F = ma_x  or  a_x = F/m

For (c) a_x = 12 N/ 2.0 kg = 6.0 m/s².

For (d) a_x = 24 N/2.0 kg = 12.0 m/s². The acceleration is directly proportional to the net force. When you double the net force, you double the acceleration.

For (e) a_x = 12 N/1.0 kg = 12.0 m/s². The acceleration is inversely proportional to the mass of the object. When you halve the mass, you double the acceleration.

1. 12 N - 4 N = 2.0 kg a_x
   4.0 m/s² = a_x
   
   
2. 24 N - 4 N = 2.0 kg a_x
   10 m/s² = a_x

It is a little difficult to solve (a) of this problem without a sneak look at the figure for part (b). There are a lot of forces here, but Fig. 3-4a below shows the only external force in the horizontal direction acting on the system of the two blocks. Again F_{net external} = ma, where a is the acceleration of the blocks and m is the mass of the system we have isolated. There are internal forces acting in the system, for example, the force of the rope on A, F_AR, the force of the rope on B, F_BR.  And we cannot forget the force of A on the rope, F_RA, or the force of B on the rope, F_RB. I have not drawn the vectors for the gravitational forces on the blocks or the normal force of the surface on the blocks because they were not asked for and there is no frictional force.



While Figures 3-4b and c above show forces  F_AR,  F_BR,  F_RA  and  F_RB,  these are internal to the system shown in Fig. 3-4a above. The only external force for the entire system in the horizontal direction is F = 12 N.

1. Thus for the system shown in Fig. 3-4a above,

   (F_{net external})_x = (mass of system)a
   12 N = (2 + 4)kg (a)  or
   12 N/6 kg = 2 m/s² = a

   For part (b), let us stare at Fig. 3-4b and c, which I reproduce below:
   
   
   Let's cope with the massless rope first:

   (F_{net external})_{on rope} = (m_R)a

   Taking to the right to be positive,

   F_RB - F_RA = (0)a, where a = the acceleration of the rope

   So,

   F_RB = F_RA

   
   It is important to notice that F_RB and F_RA, while equal and in opposite directions, are not Newton third law of motion forces because they act on the SAME object. They are equal and in the opposite direction because they must sum to zero because a massless object has no net force acting on it. Then by Newton's third law we know that:

   F_RA = - F_AR  and  F_RB = -F_BR.

   Thus since the magnitudes of these forces are equal, F_RA = F_RB, F_AR = F_BR. We throw all this careful notation to the wind and call F_AR, F_BR, F_RA and F_RB the tension T. With this information we redraw Fig. 3-4b below:
   
   
   
   
2. Now we can choose either object to find the tension T in the rope. You can check the answer by isolating the other block and finding T.
   
   

   For block B, in the X-direction,	For block A, in the X-direction,
   (Fnet external)x = (mB)a	(Fnet external)x = (mA)a
   12 N - T = (4 kg)(2 m/s2) = 8 N	T = (2 kg)(2 m/s2) = 4 N
   So  12 N - 8 N = T = 4 N

   
   
   
   

(a) For constant velocity, a = 0 and F - mg = m(0) = 0
or  F = mg = 3.0 kg (10 m/s²) = 30 N

(b) For a constant acceleration, a = 3.0 m/s², F - 30 N = 3.0 kg(3m/s²)
or  F = 9.0 N + 30 N = 39 N.

1. For entire system (Fig. 4a above),
   
   F_net = Ma
   F - 3mg = (3m)a
   9.0 N - (0.30kg)(10 m/s²) = (0.30 kg)a
   9.0 N - 3.0 N = (0.30 kg)a
   20 m/s² = a
   
   
2. To find the tension at the top of the rope T(b) first use the lower figure in Fig. 4b above.
   
   F_net = Ma
   T(b) - 2mg = (2m)a
   T(b) - 2.0 N = (0.20 kg)(20m/s²)
   T(b) - 2.0 N = 4.0 N.
   T(b) = 6.0 N
   
   Check with top figure in Fig. 4b above.
   
   F_net = Ma
   9.0 N - mg - T(b) = ma
   9.0 N -1.0 N - T(b)=(0.10 kg)(20 m/s²)=2.0 N
   6.0 N = T(b)
   
   
3. The mass of 1/5 of the rope = m/5 = 0.02 kg and its weight=mg/5= 0.2N. Use the top figure in Fig. 4c above.
   
   F_net = Ma
   9.0 N - 1.2 mg - T(c) = (0.12 kg)(20 m/s²) = 2.4 N
   T(c) = 5.4 N
   
   Using the bottom figure in Fig. 4c above,
   
   T(c) - 1.8 mg = T(c) - 1.8 N = (0.18 kg)(20 m/s²) = 3.6 N
   T(c) = 5.4 N
   
   

1. 
   
   
   

   (Fnet)x = max	(Fnet)y = may
   26 N - f = 2.0 kg(ax)	FN - mg = m(0)
   	FN = mg = (2.0 kg)(10m/s2)=20N

   
   
2.                        f = µ_kF_N = 1/5(20 N) = 4.0 N
   26 N - 4.0 N = 2.0 kg(a_x)
   
   
3. 11 m/s² = a_x

1. 
   
   
   
2. (F_net)_y = ma_y
   F_N + 26 N sin 22.6⁰ - mg = 0
   F_N = + 20 N - 26 N(5/13) = 20 N - 10 N = 10 N.
   f = µ_kF_N = (1/5)(10 N) = 2 N
   
   
3. (F_net)_x = ma_x
   26 N cos 22.6⁰ - f = 2.0 kg(a_x)
   26 N(12/13) - 2 N = 2.0 kg(a_x)
   24 N - 2 N = 2.0 kg(a_x)
   11 m/s² = a_x
   
   

(a) & (b) The x-axis is chosen in the direction of the acceleration. The axes, forces and components of forces are shown in the figure above.

(c) (F_net)_x = ma_x
30 N sin 30⁰ = 3.0 kg(a_x)
15 N = 3.0 kg(a_x)
5.0 m/s² = a_x

1. 
   
   
   
   
2. Now there is a frictional force up the plane.
   
   (F_net)_y = ma_y
   F_N - mg cos 30^o = 0
   F_N = 30 N (0.866) = 26 N
   f = µ_k F_N = 0.154(26 N) = 4.0 N
   
   
3. (F_net)_x = ma_x
   30 N sin 30^o - f = 3.0 kg(a_x)
   15 N - 4.0 N = 3.0 kg(a_x)
   a_x = 11/3m/s²
   
   

1. First isolate m₁,  taking the X-axis in the direction of the acceleration:
   
   

   (Fnet)x = m1ax	(Fnet)y = m1ay
   T - m1g sin 16.3o - f = m1ax	FN - m1g cos16.3o = m1(0)
   	FN = 25 N(24/25) = 24 N f = µkFN = 1/6(24 N) = 4 N
   T - 25 N(7/25) - 4 N = 2.5 kg a
   T - 11 N = 2.5 kg a     (Equation 1)
   Note: I have dropped the subscript on a, the only acceleration.

   
   
   Now isolate m₂,  taking down as positive:
   
   F_net = m₂a
   m₂g - T = m₂a
   20 N - T = 2.0 kg a    (Equation 2)
   
   Adding Eq. 1 and Eq. 2:

   20 N - 11 N = 4.5 kg a  or  a = 2.0 m/s²

   
   
2. Substituting a = 2.0 m/s² into Eq. 2:

   20 N - T = 2.0 kg(2.0 m/s²)  and  T = 16 N.

   
   
   

1. (F_{net external})_x = ma_x
   F_friction = mv²/r = 2 kg (2 m/s)²/(0.5 m) = 16 N
   
   
2. There is no acceleration in the Z-direction.
   (F_{net external})_z = ma_z.
   F_N - mg = m(0)  or  F_N = mg = (2 kg)(10 m/s²) = 20 N
   
   As shown in Fig. for #12b above,
   Force of table on object = {(F_friction)² + F_N²)^1/2
                                         = {(16)² + (20)²}^1/2 N
                                         = {656}^1/2 N = 25.6 N
   tan Q = F_N/F_friction = 20/16 = 1.25
   Q = 51.3^o
   
   

1. The centripetal acceleration a = v²/r = (3m/s)²/0.5 m = 18 m/s².  The object travels in a counterclockwise direction so the directions of the velocities are as shown in Fig. 7a above. Remember the velocity is always tangent to the path. Thus,
   
   
2. at B the velocity is to the right,
   
   
3. at T the velocity is to the left, and
   
   
4. at S the velocity is down.
   
   
5. At B the acceleration is into the center of the circle or up. At B we take the X-axis up.
   
   
6. At T the acceleration is into the circle or down. At T the X-axis is down.
   
   
7. At S the acceleration is to the right and the X-axis is to the right. All of the accelerations are in the direction of the net force.
   
   For all cases F_net = ma = 2 kg(18 m/s²) = 36 N.
   
   Looking at Fig. 7b below, we see that at B and T, there are no components of the force in any direction except the X-direction so we drop the subscripts on F_net.
   
   
   
   
8. At B,
   
   F_net = ma
   F_{or at B} - mg = ma
   F_{or at B} - 20 N = 36 N
   F_{or at B} = 56 N
   
   
9. At T, I have drawn F_{or at T} down, but I must use the equation for a final decision, so I write it with ± to find its sign and, therefore, its direction.
   
   F_net = ma
   ±F_{or at T} + 20 N = 36 N so
   F_{or at T} = +16 N or it is down.
   
   
10. At S,  F_{or at S}  must have a component into the center or along the +X-axis to produce the centripetal acceleration and a component along the +Y-axis so the net force along the Y-axis is zero keeping the magnitude of the velocity for the uniform circular motion constant. Thus, we must look at F_net along the X and Y-axes.
    
    
    
    

    (Fnet)x at S = max	(Fnet)y at S = may
    (For)x at S = 36 N	(For)y at S - mg = m(0) (For)y at S = mg = 20 N
    For at S = {(For)2x at S + (For)2x at S}1/2
    For at S = {(36)2 + (20)2}1/2 N = 41 N
    tan Q = (For)y/(For)x = 20 N/ 36 N = 0.55. Q = 29o

    
                                                                                                                   
    
    

1. Forces shown in Fig. 8 below:
   
   
   
   
2. (F_net)_x = ma_x         (F_net)_y = ma_y
   -mg sin Q= ma_x       T - mg cos Q= mv²/L
   
   Acceleration in the X-direction = -g sin Q
   Acceleration in the Y-direction = v²/L, where v is speed of the object at that instant and L is the radius of the circular arc.
   
   a_y = v²/L = T/m - g cos Q
   
   
3. In general, a_x = -g sin Q
      (i) at maximum displacement, a_x = -g sin Q_max
     (ii) at the equilibrium position, Q = 0, sin Q= 0, a_x = 0.
   In general, T- mg cos Q=mv²/L.
   
   
   1. At its maximum displacement, the bob comes momentarily to rest,
      v = 0 and T = mg cos Q_max.
      
      
   2. At the equilibrium position, v = v_max,
      T = mg cos 0⁰ + mv_max²/L = mg + mv_max²/L.
      
      

1. 
   
   
   The forces acting on the object are its weight mg and the tension T in the string. The acceleration is into the center or, at this instant, to the right. We take the X-axis in the direction of the acceleration and the Y-axis perpendicular to it. The weight mg is in the negative Y-direction, but T is in neither the X nor Y direction so we take components, as is shown in Figure 9 above. The acceleration in the X-direction is the centripetal acceleration v²/r. There is no acceleration in the Y-direction.
   
   
2. 
   

   (Fnet)x = max	(Fnet)y = may
   T sin Q = mv2/r      (Equation 1)	T cos Q - mg = 0  or
   	T cos Q = mg        (Equation 2)

   
   Dividing Eq. 1 by Eq. 2:

   T sin Q/T cos Q = tan Q = (mv²/r)/(mg) = v²/rg    (Equation 3)

   From geometry, we see that r/L = sin Q  or  r = L sin Q        (Equation 4)
   
   
3. Substituting Eq. 4 into Eq. 3:
   
   tan Q= v²/(L sin Q)g  or  (tan Q) (L sin Q)g= v²  and
   v =(Lg sin Qtan Q)^1/2
   
   
4. v = (distance traveled/time) = 2pr/T  or
   T= 2pr/v = 2p(L sin Q)/v =
   T = 2p(L sin Q)/(Lg sin Qtan Q)^1/2
      = 2p(L sin Q)/(Lg sin²Q/cos Q)^1/2 =
   T = 2p(L cos Q/g)^1/2
   
   

1. For motion in a horizontal circle, from the solution to #15, we see that:
   
   T sin Q= mv²/L sinQ  or
   v² = TL sin² Q/m (1)
   
   Also, T cos Q= mg  or  cos Q= mg/T
   
   From Fig. a, we see that:
   
   sin Q= (T² - m²g²)^1/2/T (2)
   
   Substituting Eq. 2 into Eq. (1):
   
   v² = TL/m (T² - m²g²)/T²
        = TL/m {1 - (mg/T)²}  or
   v = (TL/m)^1/2 {1 - (mg/T)²}^1/2
   
   
2. For a vertical circle, maximum tension occurs when the object is at the lowest point and the tension and weight of the object are in opposite directions. For this case,

   T - mg = mv²/L  or  v ={(T - mg)L/m}^1/2

   
   
   

1. Plots of the position in the x and y direction as a function of time are straight lines. The slopes of these lines are the constant velocities, v_x and v_y, respectively. For constant velocity, there is no acceleration nor force.  F_x = 0.  F_y = 0.
   
   
2. For the projectile, the object travels equal horizontal distances in equal time with a constant velocity.  F_x = 0.  There is a constant acceleration down.  F_y = - constant.
   
   
3. The speed of the particle is constant. The velocity is always tangent to the circle. This is uniform circular motion with the acceleration in toward the center. It is constant in magnitude, but changes direction.
   
   F_x = - (constant)x  and  F_y = - (constant)y.
   F = F_xi + F_yj = - (constant)(xi + yj)  or  F = - (constant)r, where r is a radial vector.
   
   
4. The slope of v_x versus t = k = a_x.
   The slope of v_y versus t = -k’ = a_y, where k and k’ are constants.
    F_x = c  and  F_y = -c’, where c and c’ are new constants.
   
   

(a) First isolate the entire system as shown in Fig. 11a above. For this system, T₂,  T₃,  the force F_mp of the platform on the man, and the force F_pm of the man on the platform are internal forces. The external forces are T₁,  the weight of the man (mg)_man, and the weight of the platform (mg)_platform.

F_{net external} = Ma
T₁ -1000 N - 600 N = (100 + 60)kg (5.0 m/s²)
T₁ = 2400 N
T₂ = T₃ = T₁/2 = 1200 N

(c) Now isolate the man as shown in Fig. 11b above:

F_{net external} = m_man a
T₃ - (mg)_man + F_mp = 100 kg (5 m/s²)
1200 N - 1000 N + F_mp = 500 N

The force of the platform on the man F_mp = 300 N.

As a check, let us isolate the platform, as in Fig. 11c above:

F_{net external} = m_platform a
T₂ - (mg)_platform - F_pm = 60 kg (5 m/s²)
1200 N - 600 N - F_mp = 300 N

Force of the man on the platform F_pm = 300 N.  F_mp and F_pm  are Newton third law of motion forces. They are equal in magnitude and opposite in direction.

Substituting Eq. 1 into Eq. 2 gives:     f_o = (1/2p)(g/µ_sR)^1/2

48 N - 12 N - 26 N = 2.0 kg(a)  or  a = 5.0 m/s²

1. Net force on sled = m_sleda_sled = 10 kg(2.0 m/s²) = 20 N = force of man on sled.
   
   
2. By Newton's third law, force of man on sled = force of sled on man
   = 20 N.
   
   
3. Net force on man = m_mana_man = 60 kg(2.0 m/s²) = 120 N = frictional force of snow on man - force of sled on man. 120 N = frictional force of snow on man - 20 N.  Frictional force of snow on man = 140 N.
   
   

1. Let F_os = the force of the scale on the object when the elevator moves with a constant velocity.  For constant velocity, F_net = ma = m(0).
   F_os - mg = 0,  and F_os = mg = (10 kg)(10 m/s²) = 100 N.  By Newton's third law, the force of the scale on the object = the force of the object on the scale. The reading of the scale is 100 N.
   
   
2. Let F'_os = the force of the scale on the object when the elevator moves upward with a constant acceleration of 5 m/s².
   F_net = ma = (10 kg)(5 m/s²).
   F'_os - mg = 50 N,  or F'_os = 50 N + 100 N = 150 N.
   F'_os = F'_so = reading of scale = 150 N.