5.1. Model: We can assume that the ring is a single massless particle in static equilibrium.

Visualize: 

Solve: Written in component form, Newton’s first law is

       

Evaluating the components of the force vectors from the free-body diagram:

      T2x = 0 N          

T1y = 0 N                  

Using Newton’s first law:

           

Rearranging:

            

Assess: Since  acts closer to the x-axis than to the y-axis, it makes sense that .

5.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables.

Visualize: 

Solve: From the lengths of the cables and the distance below the ceiling we can calculate q as follows:

Newton’s first law for this situation is

The x-component equation means . From the y-component equation:

Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension.

5.6. Visualize: Please refer to Figure Ex5.6.

Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. Applying Newton’s second law,

    

For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive x-axis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are

5.9. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force.

Visualize: 

Solve: (a) Since the box is at rest, ax = 0 m/s2, and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero.

(b) For this situation again, ax = 0 m/s2, so .

(c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ax = 5.0 m/s2,

 

Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s2.

5.11. Model: The astronaut is treated as a particle.

Solve: The mass of the astronaut is

Therefore, the weight of the astronaut on Mars is

Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than to the earth, so the smaller weight on Mars makes sense.

5.16. Model: We assume that the truck is a particle in equilibrium, and use the model of static friction.

Visualize: 

Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal force has no component in the x-direction, so we can ignore it here. For the other two forces:

Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is  of the truck’s weight seems reasonable.

5.17. Model: We assume that the car is a particle undergoing skidding, so we will use the model of kinetic friction.

Visualize: 

Solve: We begin with Newton’s second law. Although the motion is one-dimensional, we need to consider forces in both the x- and y-directions. However, we know that ay = 0 m/s2. We have

We used  because the free-body diagram tells us that  points to the left. The friction model relates  to  with the equation. The y-equation is solved to give . Thus, the kinetic friction force is . Substituting this into the x-equation yields:

The acceleration is negative because the acceleration vector points to the left as the car slows. Now we have a constant-acceleration kinematics problem. Dt isn’t known, so use

Assess: The skid marks are 136 m long. This is » 430 feet, reasonable for a car traveling at »80 mph. It is worth noting that an algebraic solution led to the m canceling out.

 

15) Explain why the coefficient of static friction is usually higher than the coefficient of kinetic friction.

When an object is sliding over another object, abrasion may occur. (See page 137 in Knight.)

5.22. Model: We will represent the bowling ball as a particle.

Visualize: 

The bowling ball falls straight down toward the earth’s surface. The bowling ball is subject to a net force that is the resultant of the weight and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion.

Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is

Since only the vertical direction matters, one can write:

When this condition is satisfied, the speed of the ball becomes the constant terminal speed . The magnitudes of the weight and drag forces acting on the ball are:

The condition for dynamic equilibrium becomes:

Assess: The value of the mass of the bowling ball obtained above seems reasonable. Such a ball is heavy so that it has a significant impact on the pins, but not “excessively” heavy so that it can be lifted and rolled by an average human player.

 

5.31. Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single vertical line: gravity and the supporting force of the elevator.

Visualize: 

Solve: (a) Before the elevator starts braking, Zach is not accelerating. His apparent weight (see Equation 5.10) is

(b) Using the definition of acceleration,

= (784 N)(1+0.340) = 1050 N

Assess: While the elevator is braking, it not only must support Zach’s weight but must also push upward on him to decelerate him, so the apparent weight is greater than his normal weight.

5.49. Model: We will model the steel cabinet as a particle. It touches the truck’s steel bed, so only the steel bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the acceleration of the truck.

Visualize: 

Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value fs max. Newton’s second law for the box and the model of static friction are

(Fnet)x = SFx = nx + wx + (fs)x = 0 N + 0 N – fs max = max = ma –fs max = ma

(Fnet)y = SFy = ny + wy + (fs)y = n - mg + 0 N = 0 N Þ n = mg

–fs max = –msn = –msmg = ma

These three equations can be combined together to get a = –msg. Because constant-acceleration kinematics gives
v12 = v02 + 2a(x1– x0) and ms = 0.8 (Table 5.1), we find

Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without making the contents slide of about 14.3 m or approximately 47 feet looks reasonable.

5.53. Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag.

Visualize: 

We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic friction  and drag  act along the –x-direction opposing the motion of the skier, the weight of the skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along the +x-direction cancel out the forces along the –x-direction.

Solve: Newton’s second law and the models of kinetic friction and drag are

(Fnet)x = SFx = nx + wx + (fk)x + (D)x = 0 N + mg sinq – fk–Av2 = max = 0 N

(Fnet)y = SFy = ny + wy + (fk)y + (D)y = n – mg cosq + 0 N + 0 N = 0 N

fk = mkn

These three equations can be combined together as follows:

(1/4)Av2 = mg sinq – fk = mg sinq – mkn  = mg sinq – mk mg cosq

Using mk = 0.06  and A = 1.8 m ´ 0.40 m  = 0.72 m2, we find

Assess: A terminal speed of 51 m/s corresponds to a speed of  This speed is reasonable but high due to the steep slope angle of 40° and a small coefficient of friction.