Home Page Textbooks Physics: Principles and Problems Page 156: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 156: Section Review

Exercise 16

Solution 1

Solution 2

Step 1

1 of 1

The force is to the center of the washing machine. The walls of the washing machine exerts a normal force (to the center) on the clothes.

Step 1

1 of 2

During the spin cycle, the drum of the washing machine moves in a circle. Where there is circular motion, the notion of centripetal force occurs. The direction of the centripetal force and acceleration is always towards the center of the circle, or in this case in the center of the tub.

Step 2

2 of 2

Since the job of the spin cycle is to remove excess water from clothes, the force is exerted by the walls of the drum. There are small notches in the wall through which water can pass.

Exercise 17

Step 1

1 of 2

begin{align*}
intertext{Part a:}
intertext{Direction of your acceleration is to the center of a circle, in moment when you are going to the right, it is to the right as well(green arrow on graph).}
intertext{Part b:}
intertext{When acceleration is to the right, your net force must be to the right too, so answer is to the right.}
intertext{Part c:}
intertext{Seat of a car.}
end{align*}\ Exercise scan

Result

2 of 2

begin{align*}
intertext{Part a: to the right}
intertext{Part b: to the right}
intertext{Part c: Seat}
end{align*}

Exercise 18

Step 1

1 of 2

The centripetal acceleration of the stone is:

$a_c = dfrac{v^2}{r} = dfrac{(2.2)^2}{0.60} = 8.0667 m/s^2$

The tension in the string is:

$$
T = m a = (0.040)*(8.0667) = 0.32 N
$$

Result

2 of 2

$$
0.32 N
$$

Exercise 19

Solution 1

Solution 2

Step 1

1 of 2

The letter to the editor should contain somewhere along the lines that $textit{textbf{there exists an acceleration due to the change in direction of the velocity}}$. Thus, there must be a $textit{textbf{net force towards the center of the circle}}$ that represents the corner turn. The road has that $textit{textbf{net force and the friction between road and the vehicle’s tires allows the aforementioned force to be exerted on the said tires}}$. The $textit{textbf{force from the car seat is exerted to the driver which accelerates them towards the center of the circle}}$. The letter should also point out that $textit{textbf{centrifugal force is not a real force}}$.

Result

2 of 2

Letter should contain a statement that explains the existence of an $textbf{acceleration due to change in direction}$, thus indicating that there exists a $textbf{net force acting with the friction}$ between the road and the tires allowing the $textbf{net force to act upon the tires}$. It should also state that $textbf{centrifugal force is not a real force}$.

Step 1

1 of 2

begin{align*}
intertext{First of all, centrifugal force is not a real force it is nonexistent force, so there is no need to balance the
centripetal and centrifugal forces. When car is turning a corner, direction of velocity is changing and because of that we have acceleration, and that acceleration has direction as change of velocity, and direction is to the center of a circle, so it is centripetal acceleration, and we also have centripetal force with the same direction. }
end{align*}

Result

2 of 2

begin{align*}
intertext{Centrifugal force is not a real force. When car is turning a corner, we have acceleration, it is centripetal acceleration, and we also have centripetal force . }
end{align*}

Exercise 20

Solution 1

Solution 2

Step 1

1 of 2

$$
a = dfrac{v^2}{r} = dfrac{2.5^2}{0.75}approx8.33
$$

$$
F = ma = 7.3times8.33 = 60.83
$$

Result

2 of 2

$$
text{color{#4257b2}60.83 N}
$$

Step 1

1 of 4

**Given:**
– Mass: $m = 7.3 mathrm{~kg}$;
– Radius: $r = 0.75 mathrm{~m}$;
– Speed: $v = 2.5 ,frac{text{m}}{text{s}}$;

**Required:**
– The force on the ball $F$;

Step 2

2 of 4

Centripetal acceleration is defined as the acceleration in uniform circular motion. Force in general is the product of mass and acceleration.
$$begin{align*}
a_text c &= frac{v^2}{r} &&(1) \
F &= ma &&(2)
end{align*}$$

Step 3

3 of 4

We are told that the ball moves around a circle, meaning that the force we have to exert on it must be equal to the centripetal force. Combining the two given equations, we have:
$$begin{align*}
F &= m a_text c \
&= frac{mv^2}{r} \
&= dfrac{7.3 mathrm{~kg} cdot left( 2.5 ,frac{text{m}}{text{s}} right)^2}{0.75 mathrm{~m}} \
&= 61 mathrm{~N}
end{align*}$$
$$boxed{ F = 61 mathrm{~N} }$$

Result

4 of 4

$$F = 61 mathrm{~N}$$

Exercise 21

Step 1

1 of 2

Earths gravity applies the force which is responsible for circular motion

Step 2

2 of 2

This motion decreases my apparent weight

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