Q3: Many forces might be acting on the object, but their vector sum (i.e., FNet = $\Sigma \$Fi) must be zero.

Q4: Of course. The important consideration is NOT whether v = 0 or not, but whether v is CHANGING. If it is not, then the net force must be zero.

Q5: If only a single force acts, then there would be no possibility of having FNet = 0. So in this case, the acceleration could not be zero.

Q7: In order to walk forward, your feet must push backward on the log, accelerating it.

P6: The average acceleration required is <a> = $\Delta \$v/ $\Delta \$t. We have vf = 0 and (changing to SI units) vi = 26.4 m/s. Thus <a> = - 3.30 m/s2. Since the mass is 950 kg, the average force required is <F> = M<a> = 3134 Newtons,

P10: ASIDE: I will adopt Giancoli's notation, as seen in Equation 4-2 on page 90: FAB will be the force ON object A exerted BY object B. Also: In this problem I will use U for the "upper box," L for the "lower box", and T for "Table." Also: Figure 4-31 on p. 104 is relevant here.

(a) In part (a) the upper box is not present (in spite of the Figure). There are two forces on the lower box: (1) an upward force FLT (the force on the lower box exerted by the table) and (2) gravity, which exerts a downward force called the "weight," given by MLg. Since the box is not accelerating vertically, these forces must balance, so

FLT = MLg = (20 kg)(9.8 m/s2) = 196 Newtons.

(b) Now the upper box is added. There are now three forces on the lower box:

(1) MLg = the weight, the downward force of the earth's gravity

(2) FLT = an upward force exerted by the table

(3) FLU = a downward force exerted by the upper box. Again, the forces must balance (we know this since the acceleration aL = 0), so (calling upward positive):

FLT - MLg - FLU = 0

We have two unknowns, (FLT and FLU) and only one equation, an embarrassing situation! But we can get a second equation by considering the forces on the upper box; there are two of these:(1) MUg = the weight, the downward force of the earth's gravity and (2) FUL = the upward force exerted by the lower box.and since, once again, the net force must be zero, we know that these must have equal magnitudes:

FUL = MUg = 98 Newtons

But we ALSO know (from the Third Law) that (ignoring direction) FUL = FLU. So, clearly, FLU (which appears in the equation for the lower box) is also equal to MUg. So our equation for the lower box becomes

FLT - MLg - MUg = 0

which can easily be solved: FLT = 294 Newtons.

P18: When the elevator (and scales and person) are accelerating, consider the forces on the person:

(1) his weight (downward force of earth's gravity) = Mg

(2) FN, the upward force of the scales (which is the only object touching him)

The net force (call upward positive) must equal Ma, so we have: $\Sigma \$F = FN - Mg = Ma

We are given that FN = (0.75)Mg, so

(0.75)Mg - Mg = Ma

which results in a = - g/4 = - 2.45 m/s2

P27: Two things touch the box: (1) the table surface, which exerts an upward force FN, and (2) the rope, which exerts an upward force FT. Gravity (the weight) is a third force, Mg, downward. Since a = 0 the net force must be zero. We are given Mg = 77.0 N.

FNet = FT + FN - Mg = 0

There are two unknowns (FN and FT), so we need a second equation. Consider the hanging mass m. There are just two forces here: (1) its weight mg downward and (2) the force of the rope upward. But the tension in the rope must be the same on the hanging mass as on the box. And since the hanging mass does not accelerate, FT = mg. Therefore, for the box:

FNet = mg + FN - Mg = 0

or

FN = Mg - mg

So (a) if mg = 30.0 Newtons, FN = 77.0 - 30.0 = 47.0 Newtons and

(b) if mg = 60.0 Newtons, FN = 17.0 Newtons

finally (c) if mg = 90 Newtons, our model gives FN = - 13 Newtons. This means that the table would have to exert a downward force of 13 Newtons if (as we have assumed) a = 0. It might do this if it were velcro, or if the box were glued down. Otherwise, our model would fail: since the upward force of the rope is greater than the box's weight, it would lift from the table, in which case, FN = 0.

P35: We are told that the FNet is parallel to the line L (call this "North"). Then the net force "East" must be zero:

FBsin(32) - FAsin(48) = 0

which gives (since we know FA = 4500 N)

FB = 6311 N.

Then

FNet = FAcos(48) + FBcos(32) = 8363 N.

P48: Comment: Figure 4-26 on page 101 shows the forces and the coordinate system used in this problem.

There is a quite good tutorial WITH FIGURES on the Sparknotes website: http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section3.rhtml

(a) The only thing touching the block is the inclined surface, so the only two forces are (1) the normal force FN, and, (2) gravity (i.e., W = Mg). Call x positive down the incline (this is convenient since we are given that the motion will be in this direction).

FN is perpendicular to the direction of motion, so it does not cause an acceleration [it is balanced by the component of the weight in the -y direction:

FN = Mgcos( $\theta \$)].

So, in the x-direction: $\Sigma \$Fx = (Mg)sin( $\theta \$) = Ma

which gives a = (g)sin(22) = 3.67 m/s2

(b) Use Equation 2-12(c) on p. 29:

v2 - (vo)2 = 2a(x - xo).

We know that $\Delta \$x = 12.0 m, a = 3.67 m/s2, and vo = 0. So

v2 = (2)(3.67)(12) = 88.1; then v = (88.1)1/2 = 9.39 m/s

P68: Comment: Figure 4-54 is a necessity here. Also: I will choose the +x direction for MA to be up the incline, the +y direction to be perpendicular to the incline toward the upper-right. Then for MB the positive direction will be downward. Consider case (a) where we know the masses and desire the acceleration and (eventually), the tension:

Mass MA has three forces on it: (1) a "normal" force FN in the +y direction exerted by the incline surface, (2) a tension FT along +x exerted by the rope, and (3) weight downward, at an angle 33 degrees relative to -y axis. Draw a Free-Body-Diagram for MA. Summing forces along the y-direction:

FN - (MA)(g)cos( $\theta \$) = 0

Make sure you can explain the zero on the RHS of the above equation!

Then, summing forces along the +x direction:

FT - (MA)(g)sin( $\theta \$) = MAa

I omit the x-subscript, since we know that the a is along the x-axis.

We have two equations, with three unknowns: FN, FT, and a. We can use the first equation to find FN, but we are not asked for this. We focus on the second equation, which contains two of our unknowns. Clearly we need another equation. We get this by considering the hanging mass, MB. There are two vertical forces, FT acting upward and (MB)g acting downward. Summing these and choosing downward as the +y direction:

(MB)(g) - FT = MBa

Now we have two equations for FT and a. The tension can easily be eliminated (by adding the two equations, for example) and we can solve for a. Then we can plug this value for a into either equation and solve for FT. The results are a = 2.23 m/s2 and FT = 7.57 Newtons.

Now, let's Consider case (b) where we know that the acceleration is zero, and the unknowns are MB and [in part (c)] the tension FT: The same equations work; only the arithmetic is different. The answers are MB = 0.545 kg and FT = 5.34 Newtons.

Comment: In case (a), where there is a non-zero acceleration, it is imperative to coordinate the choice of directions for the two masses. So since we chose the +x direction for the sliding mass to be UP the incline, we want to choose downward as the positive direction for the motion of the hanging mass. Why? Because if we do not, we cannot assume that the accelerations are the same. In fact, in that case, the acceleration of MB would be the negative of the acceleration of MA. If we want "a" in the two equations to be the same, we must make a coordinated choice. There is nothing wrong with choosing "upward" as positive in both cases, but then we must use aB = - aA in the equations for the two masses since, in this case, a positive acceleration for one necessitates a negative acceleration for the other.