Download A rightward force is applied to a book in order to move it

Normal
A rightward force is applied to a book in order to move it across
a desk at constant velocity. Consider frictional forces. Neglect air
resistance. Diagram the forces acting on the book.
FORCE
friction
(Note: Normal/Weight and friction/FORCE are the same length)
Weight
Strap Tension
1. A college student rests a backpack upon his shoulder. The pack
is suspended motionless by one strap from one shoulder. Diagram
the vertical forces acting on the backpack.
(Note: Strap Tension and Weight are the same length)
Weight
Air Resistance
2. A skydiver is descending with a constant velocity. Consider air
resistance. Diagram the forces acting upon the skydiver.
(Note: Air Resistance and Weight are the same length)
Weight
N = 12N
3. Calculate the net force in the following situation.
F=9N
fk = 3N
Net Force: ____6_______ N
Direction: ____East
W = 12N
4. What acceleration will result when a 12-N net force applied to a 3-kg object? A 6-kg object?
a) 4 m/s2
b) 2 m/s2
5. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. Determine the mass.
3.2 kg
F/m =a,
so 3F/2m = 3/2a,
so anew = 3 m/s2
6. An object is accelerating at 2 m/s2. If the net force is tripled an
doubled, then what is the
new acceleration?
80 N
8. An applied force of 50 N is used to accelerate an object
to the right across a frictional surface. The object
encounters 10 N of friction. Use the diagram to determine
the normal force, the net force, the mass, and the
acceleration of the object. (Neglect air resistance.)
8.16 kg
4.9 m/s2
40 N to right
9. An applied force of 20 N is used to accelerate an
object to the right across a frictional surface. The
object encounters 10 N of friction. Use the diagram to
determine the normal force, the net force, the
coefficient of friction ("mu") between the object and
the surface, the mass, and the acceleration of the
object. (Neglect air resistance.)
100 N
.1 N
Ffrict/Normal =
10/100
= 0.1 N
10.2 kg
.98 m/s2
10 N
Fgrav/9.8 =
100/9.8
= 10.2 kg
49 N
-4.9 N
10. A 5-kg object is sliding to the right and encountering a
friction force which slows it down. The coefficient of
friction ("mu") between the object and the surface is 0.1.
Determine the force of gravity, the normal force, the force of
friction, the net force, and the acceleration. (Neglect air
resistance.)
49 N
.98 m/s2
-4.9 N
11. A rightward force is applied to a 6-kg object to move it
across a rough surface at constant velocity. The object
encounters 15 N of frictional force. Construct a free-body
diagram to determine the gravitational force, normal force,
net force, and applied force. (Neglect air resistance.)
Normal
FORCE
15 N
friction
Weight
Normal =
W = 98 N
12. A rightward force is applied to a 10-kg object to move it
across a rough surface at constant velocity. The coefficient of
friction between the object and the surface is 0.2. Construct
a free-body diagram to determine the gravitational force,
normal force, applied force, frictional force, and net force. Friction = 
x Normal =
(Neglect air resistance.)
.2 x 98 =
19.6 N
Net Force = 0 N
FORCE =
friction = 19.6 N
Weight =
10 x 9.8 =
98 N
13. A rightward force is applied to a 5-kg object to move it across
a rough surface with a rightward acceleration of 2 m/s2. The
coefficient of friction between the object and the surface is
0.1. Construct a free-body diagram to determine the
gravitational force, normal force, applied force, frictional
Friction = 
force, and net force. (Neglect air resistance.)
x Normal =
.2 x 98 =
19.6 N
Normal =
W = 49 N
FORCE =
friction = 19.6 N
Weight = 5 x 9.8
= 49 N
Fapplied – Faccel = 25 14. A rightward force of 25 N is applied to a 4-kg object to
ma = 25 -10N = 15N
move it across a rough surface with a rightward acceleration
of 2.5 m/s/s. Use the diagram to determine the gravitational
force, normal force, frictional force, net force, and the
coefficient of friction between the object and the surface.
.383
(Neglect air resistance.)
Normal = 4kg x
9.8m/s2 = 39.2 N
Weight = 4kg x
9.8m/s2 = 39.2 N
Friction/Normal =
15/39.2 = .383
Fapplied – friction
= ma = 4 x 2.5 = 10N
1. A net force of 55N acts due west on an object. What added single force on the object produces
equilibrium?
55 N due west
55 N due east
Equilibrium means F = 0, so F1 + F2 = 0, taking west for negative, -55N + F2 = 0, F2 = +55N
2. Two forces act on an object. One force is 6.0N horizontally. The second force is 8.0N vertically.
Find the magnitude and direction of the resultant. If the object is in equilibrium, find the magnitude and
direction of the force that produces equilibrium.
82 + 62 = 10 N
8N
-1
 = tan (8/6) =
0
53.1


10 N
6N
3. A 62N force acts at 30 degrees and a second 62N force acts at 60 degrees. Determine the
resultant force. What is the magnitude and direction of the force that produces equilibrium?
Resultant =
62 sin 30 + 62 sin 60 = 84.69
119.78 N

62 cos 30 + 62 cos 60 = 84.69
Equilibrant. = 119.78 N @ 2250
4. Two forces act on an object. A 36N force acts at 225 degrees. A 48N force acts at 315 degrees.
What would be the magnitude and direction of their equilibrant?
Equilibrant. = 60 N @ 980
-(36 sin 45) + 48 sin 45 = +8.48
48 cos 45 + 36 cos 45 = 59.4

Resultant. = 60 N @ 2780
5. Two ropes are pulling on a log. One is pulling 12N at 10 degrees; the other is pulling 8N at 120
degrees. What is the net force and direction on the log?
Resultant. = 11.93 N @ 490
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