# Answers/Solutions: Giancoli Chapter 4

Q3: *Many* forces might be acting on the object, but their vector sum (i.e., **F**_{Net} = **F**_{i}) 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 F_{Net} = 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> = v/t. We have v_{f} = 0 and (changing to SI units) v_{i} = 26.4 m/s. Thus <a> = - 3.30 m/s^{2}. 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: F_{AB} 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 F_{LT} (the force on the lower box exerted by the table) and (2) gravity, which exerts a downward force called the "weight," given by M_{L}g. Since the box is not accelerating vertically, these forces must balance, so

F_{LT} = M_{L}g = (20 kg)(9.8 m/s^{2}) = 196 Newtons.

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

(1) M_{L}g = the weight, the downward force of the earth's gravity

(2) F_{LT} = an upward force exerted by the table

(3) F_{LU} = a downward force exerted by the upper box. Again, the forces must balance (we know this since the acceleration a_{L} = 0), so (calling upward positive):

F_{LT} - M_{L}g - F_{LU} = 0

We have two unknowns, (F_{LT} and F_{LU}) 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) M_{U}g = the weight, the downward force of the earth's gravity and (2) F_{UL} = 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:

F_{UL} = M_{U}g = 98 Newtons

But we ALSO know (from the Third Law) that (ignoring direction) F_{UL} = F_{LU}. So, clearly, F_{LU} (which appears in the equation for the lower box) is also equal to M_{U}g. So our equation for the lower box becomes

F_{LT} - M_{L}g - M_{U}g = 0

which can easily be solved: F_{LT} = 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) F_{N}, 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:

F = F_{N} - Mg = Ma

We are given that F_{N} = (0.75)Mg, so

(0.75)Mg - Mg = Ma

which results in a = - g/4 = - 2.45 m/s^{2}

The negative sign indicates downward acceleration.

P27: Two things touch the box: (1) the table surface, which exerts an upward force F_{N}, and (2) the rope, which exerts an upward force F_{T}. 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.

F_{Net} = F_{T} + F_{N} - Mg = 0

There are two unknowns (F_{N} and F_{T}), 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, F_{T} = mg. Therefore, for the box:

F_{Net} = mg + F_{N} - Mg = 0

or

F_{N} = Mg - mg

So (a) if mg = 30.0 Newtons, F_{N} = 77.0 - 30.0 = 47.0 Newtons and

(b) if mg = 60.0 Newtons, F_{N} = 17.0 Newtons

finally (c) if mg = 90 Newtons, our model gives F_{N} = - 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, F_{N} = 0.

P35: We are told that the **F**_{Net} is parallel to the line L (call this "North"). Then the net force "East" must be zero:

F_{B}sin(32) - F_{A}sin(48) = 0

which gives (since we know F_{A} = 4500 N)

F_{B} = 6311 N.

Then

F_{Net} = F_{A}cos(48) + F_{B}cos(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 F_{N}, 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).

F_{N} 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:

F_{N} = Mgcos()].

So, in the x-direction:

F_{x} = (Mg)sin() = Ma

which gives a = (g)sin(22) = 3.67 m/s^{2}

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

v^{2} - (v_{o})^{2} = 2a(x - x_{o}).

We know that x = 12.0 m, a = 3.67 m/s^{2}, and v_{o} = 0. So

v^{2} = (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 M_{A} to be up the incline, the +y direction to be perpendicular to the incline toward the upper-right. Then for M_{B} the positive direction will be *downward*. Consider case (a) where we know the masses and desire the acceleration and (eventually), the tension:

Mass M_{A} has three forces on it: (1) a "normal" force F_{N} in the +y direction exerted by the incline surface, (2) a tension F_{T} 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 M_{A}. Summing forces along the y-direction:

F_{N} - (M_{A})(g)cos() = 0

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

Then, summing forces along the +x direction:

F_{T} - (M_{A})(g)sin() = M_{A}a

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

We have two equations, with three unknowns: F_{N}, F_{T}, and a. We can use the first equation to find F_{N}, 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, M_{B}. There are two vertical forces, F_{T} acting upward and (M_{B})g acting downward. Summing these and choosing downward as the +y direction:

(M_{B})(g) - F_{T} = M_{B}a

Now we have two equations for F_{T} 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 F_{T}. The results are a = 2.23 m/s^{2} and F_{T} = 7.57 Newtons.

Now, let's Consider case (b) where we know that the acceleration is zero, and the unknowns are M_{B} and [in part (c)] the tension F_{T}: The same equations work; only the arithmetic is different. The answers are M_{B} = 0.545 kg and F_{T} = 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 M_{B} would be the *negative* of the acceleration of M_{A}. 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 a_{B} = - a_{A} in the equations for the two masses since, in this case, a positive acceleration for one necessitates a negative acceleration for the other.