Physics: An Introduction Chapter 6 -- Selected Solutions

Chapter 6: Applications of Newton's Laws
Selected Solutions

Selected Solutions

Solutions to selected end of chapter problems from the Walker text book.

17. (a) Determine the magnitude (always positive) of the force required to stretch the spring 2.00 cm.

F = kx = 150 N/m 0.0200 m = 3.00

Using Newton's 2^nd law we can calculate the magnitude of the force of static friction. Note that f_s,x = f_s

(b) No, the force was determined using only the spring constant and the extension of the spring.

29. (a) The weight of mass m is supported by the tension in the rope. Therefore,

T₁ = T₂ = mg = (2.50 kg)(9.81 m/s²) = 24.525 N

The force F that the leg exerts on the pulley is balanced by the net force along the central line from T₁ and T₂

F = T₁cos30^o + T₂cos30^o = 2 24.525 N cos^o = 42.5

39. (a) The x-axis is in the direction of the force. Let m₁ = 1.5 kg, m₂ = 0.93 kg.
Newton's 2^nd law for m₁ gives the equation

Newton's 2^nd law for m₂ gives the equation

Adding these two equations eliminates T to give

F = (m₁ + M₂)a

We can now solve this for the acceleration

(b) Since the tension is the only force on m₂ we can write

(c) By examining the above equation in part (b), we see that increasing m₁ would increase the denominator, and therefore decrease the tension.

47. (a) The rider moves in uniform circular motion, so the net force on the rider provides a centripetal force. Applying Newton's 2nd law at the top of the Ferris wheel (where all the forces are vertical) gives:

You can see that the normal force exerted on a rider is less than that rider's weight by an amount mv²/r, which results in an apparent weight less than the rider's actual weight.
Applying Newton's 2^nd law at the bottom of the Ferris wheel gives:

Here you can see that the normal force exerted on a rider is greater than that rider's weight by an amount mv²/r, which results in an apparent weight greater than the rider's actual weight.

(b) Let's first determine the speed at which we move around the circular path. This speed, being constant, can be calculated as distance divided by time

As discussed in part (a), the apparent weight at the top equals the normal force

The apparent weight at the bottom equals the normal force at the bottom given in part (b)

63. (a) Let T be the tension in the slanting stretch of rope. Then Tsin(45^o) is the tension in the rope supporting mass B, and Tcos(45^o is the tension in the rope pulling on mass A. But we can use the fact that sin(45^o) = cos(45^o). At the knot where the three ropes intersect we apply Newton's 2^nd law.
For the y-direction:

For the x-direction:

If mass A is truly in equilibrium we must have that

,

so the condition is satisfied.

(b) If mass A is doubled f_s,max would double, but this would not affect f_s so long as mass A is heavy enough for .