 Phyllis Fleming Physics Physics 107
 Answers - Newton's Laws
 1 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, ay = 0.  (Fnet)y = may =m(0) = 0 (a) and (b), FN - Fg = 0  or  FN = Fg = mg . For m = 2.0 kg, FN = Fg = (2.0 kg)(10 m/s2) = 20 N. (Fnet)x = max F = max  or  ax = F/m For (c) ax = 12 N/ 2.0 kg = 6.0 m/s2. For (d) ax = 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) ax = 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.
 2 Now the net force in the X-direction depends on an applied force to the right and a frictional force to the left. (Fnet)x = max F - f = max 12 N - 4 N = 2.0 kg ax 4.0 m/s2 = ax 24 N - 4 N = 2.0 kg ax 10 m/s2 = ax While the applied force F is doubled, the net force is not doubled and the acceleration is not doubled. The second net force = 20/8 x the first net force = 5/2 the first net force. The second acceleration = 5/2 the first acceleration = 5/2 x 4.0 m/s2 = 10 m/s2.

3.  In Fig. 2a above, I have isolated (shown by dashed box) the entire system in order to find the acceleration of the two blocks. In Fig. 2b above, I have isolated A and B separately in order to find FAB and FBA.  FAB is the force of B on A and FBA is the force of A on B.  FAB is an external force to the system which consists only of A and FBA is an external force to the system which consists only of B.  FAB and FBA are internal forces to the system of Fig. 2a.  I do not use a Y-axis in the figures because I am not interested in the normal forces and there is no friction in the problem. For the entire system of Fig. 2a, I know the external force that acts on it along the X-axis and I know the mass of the system so I can find its acceleration. Since I am interested only in the X-direction, I drop the subscript x on Fnet and a.

(a)  Fnet external = (mass of system)a
12 N = (2 + 4)kg (a)  or
12 N/6 kg = 2 m/s2 = a

Now we isolate block A to find FAB since we know mA and a and then isolate block B because we know everything but FBA, as shown in Fig. 2b.  F is only in contact with A and A pushes B to the right, while B pushes A to the left.

 For block A For block B (b) (Fnet external)on A = (mA)a (c) (Fnet external)on B = (mB)a F - FAB = (2 kg)(2 m/s2) FBA = (4 kg)(2 m/s2) 12 N - FAB = 4 N FBA = 8 N or 12 N - 4 N = 8 N = FAB

FAB and FBA are Newton third law of motion forces. They are equal in magnitude, but opposite in direction.  FAB, the force of B on A, has a magnitude of 8 N and is to the left.  FBA, the force of A on B, has a magnitude of 8 N and is to the right.

4.

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 Fnet 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, FAR, the force of the rope on B, FBR.  And we cannot forget the force of A on the rope, FRA, or the force of B on the rope, FRB. 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  FAR,  FBR,  FRA  and  FRB,  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,
(Fnet 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:
(Fnet external)on rope = (mR)a
Taking to the right to be positive,
FRB - FRA = (0)a, where a = the acceleration of the rope
So,
FRB = FRA

It is important to notice that FRB and FRA, 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:
FRA = - FAR  and  FRB = -FBR.
Thus since the magnitudes of these forces are equal, FRA = FRB, FAR = FBR. We throw all this careful notation to the wind and call FAR, FBR, FRA and FRB 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

 5 The forces acting on the object are the applied force F and the weight of the object mg. In all cases, Fnet = ma (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.

6.  1. For entire system (Fig. 4a above),

Fnet = 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.

Fnet = 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.

Fnet = 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.

Fnet = 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

7.

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 = µkFN = 1/5(20 N) = 4.0 N
26 N - 4.0 N = 2.0 kg(ax)

3. 11 m/s2 = ax
 8 (Fnet)y = may FN + 26 N sin 22.60 - mg = 0 FN = + 20 N - 26 N(5/13) = 20 N - 10 N = 10 N. f = µkFN = (1/5)(10 N) = 2 N (Fnet)x = max 26 N cos 22.60 - f = 2.0 kg(ax) 26 N(12/13) - 2 N = 2.0 kg(ax) 24 N - 2 N = 2.0 kg(ax) 11 m/s2 = ax
 9 (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) (Fnet)x = max 30 N sin 300 = 3.0 kg(ax) 15 N = 3.0 kg(ax) 5.0 m/s2 = ax
 10 Now there is a frictional force up the plane. (Fnet)y = may FN - mg cos 30o = 0 FN = 30 N (0.866) = 26 N f = µk FN = 0.154(26 N) = 4.0 N (Fnet)x = max 30 N sin 30o - f = 3.0 kg(ax) 15 N - 4.0 N = 3.0 kg(ax) ax = 11/3m/s2

11. 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:

Fnet = 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/s2and  T = 16 N.

 12 Figure for #12a above is a sketch of the object moving on the table and all of the forces acting on it. As the object starts to slip away from the circle, a frictional force acts into the center of the circle to produce the centripetal acceleration. I chose the X-axis to the right because at the moment shown in the picture, the centripetal acceleration is into the center of the circle or to the right. The frictional force is the only force into the circle or, as I have drawn it, along the positive X-axis. This is a three dimensional drawing showing that for clockwise motion, at this instant of time, the velocity is along the positive Y-axis. The weight of the object is always along the negative Z-axis and the normal force of the table along the positive Z-axis. Now our axis of interests are the X-axis and the Z-axis. The acceleration in the positive X-direction is the centripetal acceleration = v2/r.                                     (Fnet external)x = max Ffriction = mv2/r = 2 kg (2 m/s)2/(0.5 m) = 16 N There is no acceleration in the Z-direction. (Fnet external)z = maz. FN - mg = m(0)  or  FN = mg = (2 kg)(10 m/s2) = 20 N As shown in Fig. for #12b above, Force of table on object = {(Ffriction)2 + FN2)1/2                                       = {(16)2 + (20)2}1/2 N                                       = {656}1/2 N = 25.6 N tan Θ = FN/Ffriction = 20/16 = 1.25 Θ = 51.3o

13. 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 Fnet = 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 Fnet. 8. At B,

Fnet = ma
For at B - mg = ma
For at B - 20 N = 36 N
For at B = 56 N

9. At T, I have drawn For 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.

Fnet = ma
±For at T + 20 N = 36 N so
For at T = +16 N or it is down.

10. At S,  For 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 Fnet 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 Θ = (For)y/(For)x = 20 N/ 36 N = 0.55. Θ = 29o

 14 Forces shown in Fig. 8 below: (Fnet)x = max         (Fnet)y = may -mg sin Θ= max       T - mg cos Θ= mv2/L Acceleration in the X-direction = -g sin Θ 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. ay = v2/L = T/m - g cos Θ In general, ax = -g sin Θ    (i) at maximum displacement, ax = -g sin Θmax   (ii) at the equilibrium position, Θ = 0, sin Θ= 0, ax = 0. In general, T- mg cos Θ=mv2/L. At its maximum displacement, the bob comes momentarily to rest, v = 0 and T = mg cos Θmax. At the equilibrium position, v = vmax, T = mg cos 00 + mvmax2/L = mg + mvmax2/L.

15.

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 Θ = mv2/r      (Equation 1) T cos Θ - mg = 0  or T cos Θ = mg        (Equation 2)

Dividing Eq. 1 by Eq. 2:
T sin Θ/T cos Θ = tan Θ = (mv2/r)/(mg) = v2/rg    (Equation 3)
From geometry, we see that r/L = sin Θ  or  r = L sin Θ        (Equation 4)

3. Substituting Eq. 4 into Eq. 3:

tan Θ= v2/(L sin Θ)g  or  (tan Θ) (L sin Θ)g= v2  and
v =(Lg sin Θtan Θ)1/2

4. v = (distance traveled/time) = 2πr/T  or
T= 2πr/v = 2 π(L sin Θ)/v =
T = 2 π(L sin Θ)/(Lg sin Θtan Θ)1/2
= 2 π(L sin Θ)/(Lg sin2Θ/cos Θ)1/2 =
T = 2 π(L cos Θ/g)1/2

 16 For motion in a horizontal circle, from the solution to #15, we see that: T sin Θ= mv2/L sinΘ  or v2 = TL sin2 Θ/m (1) Also, T cos Θ= mg  or  cos Θ= mg/T From Fig. a, we see that: sin Θ= (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 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
 17 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, vx and vy, respectively. For constant velocity, there is no acceleration nor force.  Fx = 0.  Fy = 0. For the projectile, the object travels equal horizontal distances in equal time with a constant velocity.  Fx = 0.  There is a constant acceleration down.  Fy = - constant. 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. Fx = - (constant)x  and  Fy = - (constant)y. F = Fxi + Fyj = - (constant)(xi + yj)  or  F = - (constant)r, where r is a radial vector. The slope of vx versus t = k = ax. The slope of vy versus t = -k’ = ay, where k and k’ are constants.  Fx = c  and  Fy = -c’, where c and c’ are new constants.

18.

(b) (a) First isolate the entire system as shown in Fig. 11a above. For this system, T2,  T3,  the force Fmp of the platform on the man, and the force Fpm 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.

Fnet 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:

Fnet external = mman a
T3 - (mg)man + Fmp = 100 kg (5 m/s2)
1200 N - 1000 N + Fmp = 500 N

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

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

Fnet external = mplatform a
T2 - (mg)platform - Fpm = 60 kg (5 m/s2)
1200 N - 600 N - Fmp = 300 N

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

19. First isolate the entire system (Fig. 12a above). For the entire system, the external forces are the applied force F, the frictional force f2 of the surface on m2,  the normal force FN on 1 & 2 of the surface on the two blocks and the attraction of the earth for the objects or the weight of the objects, (m1 + m2)g.

As the objects move to the right,  f2 acts to the left.

 (Fnet)x = Max (Fnet)y = May 45 N - f2 = 6.0 kg a FN on 1 & 2 - 60 N = 0 FN on 1 & 2 = 60 N f2 = µk FN on 1 & 2  = 0.25(60 N) = 15 N 45 N - 15 N = 6.0 kg a 5.0 m/s2 = a

Now isolate m1 (Fig. 12b above). There must be a frictional force  f1  on  m1  due to  m2  to the right to produce the acceleration of  m1.

 (Fnet)x = m1ax (Fnet)y = m1ay f1 = 2.0 kg(5 m/s2) FN1 - 20 N = 0 FN1 = 20 N  and  f1 = µs(20 N) µs(20 N) = 10 N µs = 1/2

As a check isolate  m2  (Fig. 12c above). By Newton's third law, m1 must exert a force  f1  on  m2  to the left.

(Fnet)x = m2ax
45 N - f2 - f1 = 4.0 kg(5 m/s2)  or
45 N - 15 N - f1 = 20 N

and again,

f1 =10 N  with  µs(20 N) = 10 N  and
µs = 1/2.

20. As the block starts to slip down, a frictional force f opposes the motion.

 (Fnet)x = max (Fnet)y = may N = mv2/R f - mg = 0  or  f = mg = µsN Since v = 2 πR/To = 2 πRfo N = mg/µs        (Equation 1) N = m(4 π2R2fo2)/R (Equation 2)

Substituting Eq. 1 into Eq. 2 gives:     fo = (1/2 π)(g/µsR)1/2

 21 Take the X-axis in the direction of the acceleration or up the plane and the Y-axis perpendicular to it  (Fig. 14 above). (Fnet)x max  F cos 37o - mg sin 37o - f = ma 60 N(4/5) - 20 N(3/5) - f = 2.0 kg(a) 48 N - 12 N - f = 0                 (Equation 1) (Fnet)y = may FN - F sin 37o - mg cos 37o = m(0) FN = 60 N(3/5) + 20 N(4/5) = 52 N f = µkFN = 1/2(52 N) = 26 N     (Equation 2) Substituting Eq. 2 into Eq. 1: 48 N - 12 N - 26 N = 2.0 kg(a)  or  a = 5.0 m/s2
 22 Net force on sled = msledasled = 10 kg(2.0 m/s2) = 20 N = force of man on sled. By Newton's third law, force of man on sled = force of sled on man = 20 N. Net force on man = mmanaman = 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.
 23 Let Fos = the force of the scale on the object when the elevator moves with a constant velocity.  For constant velocity, Fnet = ma = m(0). Fos - mg = 0,  and Fos = 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. Let F'os = the force of the scale on the object when the elevator moves upward with a constant acceleration of 5 m/s2. Fnet = 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.
 24 Fnet = mg - f = ma or a = g - f/m. The lead sphere has a greater density than the wood sphere. The volume of the spheres = 4/3 πr3 and the mass m of the spheres = density x volume. The lead sphere has the same volume as the wood sphere since their radii are identical, but it has a larger mass because it has a greater density. Since a = g - f/m, the sphere of the larger mass has the greater acceleration.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 April 17, 2003