Home Page Textbooks Physics: Principles and Problems Page 135: Section Review

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 135: Section Review

Exercise 42

Step 1

1 of 2

This method is very useful if the rope is inextensible. As shown in the force diagram, a small vertical force on the rope causes a large force in the horizontal direction. The force of tension acts mostly in the horizontal direction, which means that it must be large in order for its vertical component to cancel the external force with which we act on the rope. Exercise scan

Result

2 of 2

The tension force has a relatively small vertical component so it must be large to cancel out the relatively small external push force.

Exercise 43

Step 1

1 of 3

Step 2

2 of 3

The diagram shows the situation for only a fraction of the weight held by one cable. This only serves to make it easier for us to determine which component cancels the weight.

$$
begin{align*}
F_T&=1300 mathrm{N} \
theta_6&=8text{textdegree} \
theta_4&=10text{textdegree} \
m&=? \
\
F_G&=6cdot F_Tcdot cos(theta_6cdot ) + 4cdot F_Tcdot cos(theta_4) \
mcdot g&=6cdot F_Tcdot cos(theta_6cdot ) + 4cdot F_Tcdot cos(theta_4) \
m&=dfrac{ F_Tcdot (6cdot cos( theta_6cdot ) + 4cdot cos(theta_4))}{g} \
&=dfrac{1300mathrm{N}cdot (6cdot cos( 8text{textdegree} ) + 4cdot cos(10text{textdegree}))}{9.8text{textdegree}}\
&=1311 mathrm{kg}
end{align*}
$$

Result

3 of 3

$$
m=1311 mathrm{kg}
$$

Exercise 44

Step 1

1 of 2

The skier is acted upon by pulling force in one direction and a component of gravity parallel to the slope in the other direction. Let the positive direction be in the direction in which the skier is pulled.

$$
begin{align*}
m&=63 mathrm{kg} \
theta&=14text{textdegree} \
F_p&=512 mathrm{N} \
mu&=0.27 \
a&= ?
\
F&=F_p-F_f-F_gcdot sin(theta) \
mcdot a&=F_p-mucdot mcdot g-mcdot gcdot sin(theta) \
a&=dfrac{F_p-mucdot mcdot g-mcdot gcdot sin(theta)}{m} \
&=dfrac{512 mathrm{N} – 0.27cdot 63 mathrm{kg}cdot cos(14text{textdegree}) – 63 mathrm{kg}cdot 9.8 mathrm{m/s^2}cdot sin(14text{textdegree})}{63 mathrm{kg}} \
&=3.189 mathrm{m/s^2}
end{align*}
$$

Result

2 of 2

$a=3.189 mathrm{m/s^2}$ up the slope.

Exercise 45

Step 1

1 of 3

Step 2

2 of 3

The diagram shows the forces on the right side of the painting. We see that $F_g = cos (theta) cdot F_T$ holds.
From the above expression we see that the left side is the maximum for the minimum angle $theta$. So the smaller the angle $theta$, the larger the horizontal component of the tension force $cos(theta)cdot F_T$.

A better way of hanging is shown in Figure b.

Result

3 of 3

The painting should be hung as shown in 5-15b.

Exercise 46

Step 1

1 of 2

The scenario described is impossible. For this situation to be possible the coefficient of friction should be negative which is impossible. In addition, the force of friction should cause motion and we know that it is just the opposite and that it always acts opposite to the direction of motion.

Result

2 of 2

No.

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