Answers/Solutions: Giancoli Chapter 5

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Q1: The only thing touching the crate is the bed of the truck, so the only forces possible are gravity and the force of the truck bed. Gravity acts downward, so it has no part in accelerating the crate horizontally. The truck bed exerts a normal force upward, but this force also has no influence on the crate's horizontal motion. But the truck bed also exerts a frictional force on the crate parallel to the bed's surface (and thus horizontally) and it is force of static friction which causes the crate to accelerate and not slide on the truck bed when the truck accelerates. Comment: If there were no friction, the truck would simply "drive out from under" the crate, and it would fall down to the ground. If the acceleration is so great that the maximum static value cannot provide the necessary force for the truck's acceleration (i.e., \mu \ sFN < MCaT), the crate slides.




Q10: Yes. The centripetal acceleration is v2/R. Increasing v while leaving R unchanged does indeed increase the acceleration. In fact, an increase of v from 50 units to 70 units (with R the same) just about doubles the acceleration!




Q17: The acceleration is toward the pole, at right angles to it, along the radius of its circular path.




P13: The horizontal force of the car on the ground must (by the Third Law) equal that of the ground on the car. This force, then, must accelerate horizontally the car+trailer. So

a = (3600 Newtons)/(1280 kg + 350 kg) = 2.21 m/s2

Now, knowing the acceleration of the trailer, the NET force on it, must be

FNet = (350 kg)(2.21 m/s2) = 773 Newtons.

There are two forces acting horizontally on the trailer: (1) FTC, the force on the trailer from the car and (2) kinetic friction. The force of kinetic friction is Ffr = \mu \ kFN. Since the trailer is on flat ground, FN = mg = (350 kg)(9.8 m/s2) = 3430 Newtons, and then Ffr = (0.15)(3430) = 514.5 Newtons. Finally, since

FNet = FTC - Ffr

we have

773 = FTC - 514.5

or: FTC = 1288 Newtons




P18: This is exactly the same physical situation as P48 in Chapter 4 except that now there is friction. We set up the problem in exactly the same way but there is an additional force: the force of kinetic friction. This force, Ffr, acts in a direction up the incline (because we know that the motion will be down the incline). We choose +x down the incline. The forces along the coordinate axes are then:

Along x-axis:

+ (Mg)sin(\theta \ ), a component of the weight (due to earth's gravity)

- Ffr = - \mu \ kN, the force of kinetic friction parallel to incline surface

Along y-axis: + FN, the "normal" force perpendicular to the incline surface

- (Mg)cos(\theta \ ), a component of the weight (due to earth's gravity)

The Equations of Motion (from Newton's Second Law) are, then:

\Sigma \ Fx = (Mg)sin(\theta \ ) - \mu \ kFN = Max = Ma

\Sigma \ Fy = FN - (Mg)cos(\theta \ ) = May = 0

(a) We can immediately solve the second of these for FN [ = (Mg)cos(\theta \ )], then substitute that result into the first and solve for a. The result is:

a = g[sin(\theta \ ) - \mu \ kcos(\theta \ ) = 2.454 m/s2

(b) Use Equation 2-12(c) on page 29 again: we know vo ( = 0), a, and \Delta \ x. The result is that v = 6.324 m/s




P30: Since we are told that MB descends, we choose our coordinate system so that +x for MA is up the incline, +y for MA upward perpendicular to the incline surface, and +y for MB is downward. (Again: it is important to coordinate the choices. If MB moves down, then clearly MA must slide up the incline. The accelerations will be the same so long as we choose positive directions for the axes to be along the direction of the motions.) The forces along the coordinate axes are:

On MA along its x-axis:

- (MAg)sin(\theta \ ), a component of the weight (due to earth's gravity)

- Ffr = - \mu \ kFN, the force of kinetic friction parallel to incline surface. (The friction will act downward along the incline since the mass slides upward.)

+ FT, the tension in the rope

On MA along its y-axis:

+ FN, the "normal" force perpendicular to the incline surface

- (MAg)cos(\theta \ ), a component of the weight (due to earth's gravity)

On MB along its y-axis:

- FT, the tension in the rope

+ MBg, its weight

So we get three equations of motion from Newton's Second Law:

\Sigma \ (FA)x: FT - (MAg)sin(\theta \ ) - \mu \ kFN = MAa

\Sigma \ (FA)y: FN - (MAg)cos(\theta \ ) = 0

\Sigma \ (FB)y: MBg - FT = MBa

(a) Three equations, the unknowns are a, FN, and FT. We can immediately use the third equation to obtain an expression for FT and the second to obtain an expression for FN. Substitute these into the first equation and solve for a. The result is a = 1.6 m/s2

(b) Use the same three equations (these constitute our mathematical model), but now we know that the acceleration a = 0 but we do not know \mu \ k. The math is even simpler and the answer is \mu \ k = 0.53.




P34: Figure 5-23 on page 126 is important for this problem, which is similar to Example 5-14.

Only one horizontal force acts: the force of static friction (assuming no skidding) between the tires and the roadbed. This force provides the necessary "centripetal" force to cause the car to move in a circular path:

Ffr = Mv2/R

The maximum v would require the maximum static frictional force, which we know is

(Ffr)Max = \mu \ SFN

where FN is the "normal" force perpendicular to the roadbed. We can easily find FN in this problem by considering the vertical forces on the car (see Figure 5-23). Since there is no vertical acceleration, these two forces must balance: FN = Mg.

So the condition for maximum speed is

\mu \ SMg = Mv2/R

The mass divides out, and we obtain (solving for v)

v = sqrt[\mu \ SRg] = 22.6 m/s (which equals 81 km/hr or 50 miles/hour)




P44: (a) Envision the bucket moving in a vertical circle at the instant when it is at the lowest point. Two forces act: gravity (downward) and the tension in the rope (upward). The vector sum (choosing the positive direction as toward the circle's center so that the acceleration will be positive) must equal the centripetal force:

FT - Mg = Mv2/R

Since we are given M, R, and FT, we can easily solve for v. The result is v = 1.72 m/s.

(b) Now the bucket is momentarily at its highest point. Both the tension in the rope and the bucket's weight act downward (which is positive - toward the center of the circular path). So we have:

FT + Mg = Mv2/R

But, at the critical speed (below which the rope would indeed "go slack"), FT = 0. So at this speed

Mg = Mv2/R

Now the mass divides out and v = sqrt(Rg) = 3.28 m/s




P58: We look at a version of Example 15 (page 127) and Figure 5-24 but with the addition of a frictional force parallel to the roadbed. Now our vehicle may be going too fast for the horizontal component of the normal force to provide the needed centripetal force to keep the car moving in a circle (depending on the given speed).

From Example 15, we see that the "proper banking angle" for a speed of 65 km/hr (= 18.1 m/s) and a turn radius of 85 meters will be Arctan(v2/Rg) = 21.4 degrees. For this angle and radius, a car moving at 65 km/hr will not skid even if hitting an oil slick (i.e., if friction = zero)!

What if the car is moving at 95 km/hr (= 26.4 m/s)? Clearly at this higher speed it WILL skid if there is no help from friction to provide additional centripetal force! We use a mathematical model identical to that of Example 5-15 but with the addition of friction between the tires and roadbed.

Draw a sketch of Figure 5-24, but with a frictional force Ffr pointing "down & to the right" along the inclined roadbed. Resolving this vector into components, we have:

(Ffr)x = Ffrcos(\theta \ )

(Ffr)y = - Ffrsin(\theta \ )

Now the equations of motion from Newton's Second Law become:

\Sigma \ Fx = FNsin(\theta \ ) + Ffrcos(\theta \ ) = Mv2/R

\Sigma \ Fy = FNcos(\theta \ ) - Ffrsin(\theta \ ) - Mg = 0

Also, we know that the maximum value of the static friction (which we will need at the maximum speed) is Ffr = \mu \ SFN. Since we are interested in the case where 95 km/hr (= 26.4 m/s) is the maximum speed, we substitute \mu \ SFN for Ffr in these two equations.

So our unknowns are \theta \ and FN. We can solve the second equation for FN:

FN = Mg/[cos(\theta \ ) - \mu \ Ssin{\theta \ ]

Then we must substitute this expression into the first equation and solve for \mu \ S. When this is done (the algebra is a little messy, but not too bad) the answer becomes

\mu \ S = 0.335

This problem is too long for a test, but I might well give you a simpler variant of the same physical situation to test your understanding of the principles.