Newton Laws of Motion

1. Problem A 31
NAME ______________________________________ DATE _______________ CLASS ____________________
Forces and the Laws of Motion
Problem A
DRAWING FREE-BODY DIAGRAMS
P R O B L E M
In the early morning, a park ranger in a canoe is observing wildlife on
the nearby shore. The Earth’s gravitational force on the ranger is 760 N
downward and its gravitational force on the boat is 190 N downward.
The water keeps the canoe afloat by exerting a 950 N force upward on it.
Draw a free-body diagram of the canoe.
S O L U T I O N
1. Identify the forces acting on the object and the directions of the forces.
• The Earth exerts a force of 190 N downward on the canoe.
• The park ranger exerts a force of 760 N downward on the canoe.
• The water exerts an upward force of 950 N on the canoe.
2. Draw a diagram to represent the isolated object.
The canoe can be represented by a simple outline, as shown in (a).
3. Draw and label vector arrows for all external forces acting on the object.
A free-body diagram of the canoe will show all the forces acting on the canoe
as if the forces are acting on the center of the canoe. First, draw and label the
gravitational force acting on the canoe, which is directed toward the center
of Earth, as shown in (b). Be sure that the length of the arrow approximately
represents the magnitude of the force.
Next, draw and label the downward force that is exerted on the boat by the
Earth’s gravitational attraction on the ranger, as shown in (c). Finally, draw
and label the upward force exerted by the water on the canoe as shown in (d).
Diagram (d) is the completed free-body diagram of the floating canoe.
Copyright
©
by
Holt,
Rinehart
and
Winston.
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rights
reserved.
(a)
(b)
(c) (d)
FEarth-on-canoe
FEarth-on-canoe FEarth-on-canoe
Fwater-on-canoe
Franger-on-canoe Franger-on-canoe
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2. Holt Physics Problem Workbook
32
NAME ______________________________________ DATE _______________ CLASS ____________________
1. After a skydiver jumps from a plane, the only force initially acting on
the diver is Earth’s gravitational attraction. After about ten seconds of
falling, air resistance on the diver will have increased so that its magni-
tude on the diver is now equal in magnitude to Earth’s gravitational force
on the diver. At this time, a diver in a belly-down position will be falling
at a constant speed of about 190 km/h.
a. Draw a free-body diagram of the skydiver when the diver initially
leaves the plane.
b. Draw a free-body diagram of the skydiver at the tenth second of
the falling.
2. A chef places an open sack of flour on a kitchen scale. The scale reading
of 40 N indicates that the scale is exerting an upward force of 40 N on
the sack. The magnitude of this force equals the magnitude of the force
of Earth’s gravitational attraction on the sack. The chef then exerts an
upward force of 10 N on the bag and the scale reading falls to 30 N.
Draw a free-body diagram of the latter situation.
3. A music box within the toy shown below plays tunes when the toy is
pushed along the floor. As a child pushes along the handlebars with
a force of 5 N, the floor exerts a force of 13 N directly upward on the toy.
The Earth’s gravitational force on the toy is 10 N downward while inter-
actions between the wheels and the floor produce a backward force of
2 N on the toy as it moves. Draw a free-body diagram of the toy as it is
being pushed.
Copyright
©
by
Holt,
Rinehart
and
Winston.
All
rights
reserved.
ADDITIONAL PRACTICE
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3. Section Two — Problem Workbook Solutions II Ch. 4–1
Forces and the
Laws of Motion 4
Problem Workbook Solutions
II
Copyright
©
by
Holt,
Rinehart
and
Winston.
All
rights
reserved.
Additional Practice A
Givens Solutions
1. a. b.
FEarth-on diver
Fair resistance-on diver
FEarth-on diver
2.
Fchef-on-sack
Fscale-on-sack
FEarth-on-sack
3.
Ffloor-on-toy-vertical
Ffloor-on-toy-horizontal
Fhandlebars-on-toy
FEarth-on-toy
1. mw = 75 kg
mp = 275 kg
g = 9.81 m/s2
The normal force exerted by the platform on the weight lifter’s feet is equal to and
opposite of the combined weight of the weightlifter and the pumpkin.
Fnet = Fn − mwg − mpg = 0
Fn = (mw + mp)g = (75 kg + 275 kg) (9.81 m/s2
)
Fn = (3.50 × 102
kg)(9.81 m/s2
) = 3.43 × 103
N
Fn = 3.43 × 103
N upward against feet
Additional Practice B
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