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Serway and Vuille - College Physics (INT) 9/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 24 / 28

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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EXAMPLE 4.7 Sled on a Frictionless Hill
(a) A sled tied to a tree on a friction less hill. (b) A diagram of forces acting on the sled.
Goal Use the second law and the normal force in an equilibrium problem.

Problem A sled is tied to a tree on a frictionless, snow-covered hill, as shown in figure (a). If the sled weighs 77.0 N, find the magnitude of the tension force T with arrow exerted by the rope on the sled and that of the normal force n with arrow exerted by the hill on the sled.

Strategy When an object is on a slope, it's convenient to use tilted coordinates, as in figure (b), so that the normal force n with arrow is in the y-direction and the tension force T with arrow is in the x-direction. In the absence of friction, the hill exerts no force on the sled in the x-direction. Because the sled is at rest, the conditions for equilibrium, ΣFx = 0 and ΣFy = 0 apply, giving two equations for the two unknowns—the tension and the normal force.
SOLUTION
Apply Newton's second law to the sled, with a with arrow = 0.
F with arrow = T with arrow + n with arrow + F with arrowg = 0
Extract the x-component from this equation to find T. The x-component of the normal force is zero, and the sled's weight is given by mg = 77.0 N.
Fx		 =T + 0 − mg sin θ = T − (77.0 N)(sin 30.0°) = 0
T = 38.5 N
Write the y-component of Newton's second law. The y-component of the tension is zero, so this equation will give the normal force.
Fy		 =0 + n − mg cos θ = n − (77.0 N)(cos 30.0°) = 0
n = 66.7 N
LEARN MORE
Remarks Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled is on the slope. This is because only part of the force of gravity (the x-component) is acting to pull the sled down the slope. The y-component of the force of gravity balances the normal force.

Question Consider the same scenario on a hill with a steeper slope.
Would the magnitude of the tension in the rope get larger, smaller, or remain the same as before?
    

How would the normal force be affected?
    
PRACTICE IT
Use the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless, snow-covered hill, as shown in Figure (a). If the sled weighs 75.0 N, find the magnitude of the tension force T with arrow exerted by the rope on the sled and that of the normal force n with arrow exerted by the hill on the sled.
T = N
n = N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs onto the sled. If the tension force is measured to be 59.5 N, find the weight of the child and the magnitude of the normal force acting on the sled.
w = N
n = N
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Active Figure 4.20 The Atwood Machine

Instructions Click 'start' to utilize the following Active Figure.

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When two objects of unequal masses are suspended on opposite sides of a frictionless pulley with negligible mass, the arrangement is called an Atwood machine. Suppose an Atwood machine has a mass of m1 = 14.6 kg and another mass of m2 = 16.1 kg hanging on opposite sides of the pulley. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord.

Conceptualize
Imagine the situation pictured in the Active Figure. As one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude.

Categorize
The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the string connected to them. Therefore, we can categorize this problem as one involving two particles under a net force.

Analyze
The free-body diagrams for the two objects are shown in the Active Figure. Two forces act on each object: the upward force T with arrow exerted by the string and the downward gravitational force. In problems such as this one the pulley is modeled as massless and frictionless, and the tension in the string is the same on both sides of the pulley. If the pulley has mass or is subject to friction, other techniques must be used to take these factors into account.

The signs used in problems such as this require care. When object 1 accelerates upward, object 2 accelerates downward. Therefore, m1 going up and m2 going down should be represented equivalently by the same acceleration with the same sign. We can do that by defining our sign convention with up as positive for m1 and down as positive for m2.

With this sign convention, the y-component of the net force exerted on object 1 is T − m1g, and the y-component of the net force exerted on object 2 is m2g − T.

Applying Newton's second law to m1 gives the result
(1)
Fy
= T − m1g = m1ay.
Do the same thing for m2:
(2)
Fy
 = m2g − T = m2ay.
The acceleration ay is the same for both objects. When Equation (1) is added to Equation (2), T cancels and we obtain
(3) −m1g + m2g = m1ay + m2ay.
Solving Equation (3) for ay and substituting for the given values of m1 = 14.6 kg and m2 = 16.1 kg gives
(4)
ay = 
m2 − m1
m1 + m2
 g = m/s2.
Substitute Equation (4) into Equation (1) to find T.
T = m1(g + ay) = 
2m1m2
m1 + m2
 g = N.
Finalize
The acceleration given by Equation (4) can be interpreted as the ratio of the magnitude of the unbalanced force (m2 − m1)g to the total mass (m1 + m2) of the system, as expected from Newton's second law. Notice that the sign of the acceleration depends on the relative masses of the two objects.
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A 4.8-g bullet leaves the muzzle of a rifle with a speed of 300 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.93-m-long barrel of the rifle?
N
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In a tug-of-war between two athletes, each pulls on the rope with a force of 270 N. What is the tension in the rope?
N

If the rope doesn't move, what horizontal force does each athlete exert against the ground?
N
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If an object is in equilibrium, which of the following statements is not true?
    
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An object of mass m is dropped from the roof of a building of height h. While the object is falling, a wind blowing parallel to the face of the building exerts a horizontal constant force F on the object.
(a) How long does it take the object to strike the ground? Express the time t in terms of g and h.
t =


(b) Find an expression in terms of m and F for the acceleration ax of the object in the horizontal direction (taken as the positive x-direction).
ax =


(c) How far is the object displaced horizontally before hitting the ground? Answer in terms of m, F, g, and h.
Δx =


(d) Find the magnitude of the object's acceleration while it is falling, using the variables F, m, and g.
a =
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Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 46 N acts as shown on the 4.0-kg block.
(a) Determine the acceleration given this system.
m/s2 (to the right)

(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks.
N

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.
N

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A block of mass m = 5.4 kg is pulled up a θ = 23° incline as in the figure with a force of magnitude F = 38 N.
(a) Find the acceleration of the block if the incline is frictionless.
m/s2

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.
m/s2
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How many significant figures are there in each of the following?
(a)    7.9 ± 0.2


(b)    3.34 ± 0.01


(c)    2.14 × 10−6


(d)    0.020180
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 1.1: SIGNIFICANT FIGURES

  • Part 1 - Multiplication and Division

    Dividing 2.54 by 1.3, a calculator produces 1.9538462. However, it would be very imprecise to report 1.9538462 as the answer because it would be leading others astray since this many significant figures in the result implies that the data used is accurate to at least that same number of significant figures. The least accurate number, 1.3, is known to two significant figures; therefore, the answer should be rounded to two significant figures, 2.0, where 1.95 is rounded up to 2.0.

    A general rule is that your answer from the product or division of two numbers should never have more significant figures than the number possessing the least number of significant figures.

    VIDEO EXAMPLE 1:    EXAMPLE 1-1

    Note that 0.0000356 has only three significant figures. The four zeroes between the decimal point and the 3 show only the position of the decimal point and are not considered significant.


Carry out the following arithmetic operations. (Enter your answers to the correct number of significant figures.)
(a) the sum of the measured values 801, 39.6, 0.50, and 5.5
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(b) the product 1.2 × 2.647
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(c) the product 5.88 × π
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