Chapter 3 P20 Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 3 P20. The helicopter view in Figure P3.20 shows two people pulling on a stubborn mule.
Find
(a) the single force that is equivalent to the two forces
shown and
(b) the force a third person would have to exert on the mule
to make the net force equal to zero.
The forces are measured in units of newtons (N).
Page 1 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 3 P20. The helicopter view in Figure P3.20 shows two people pulling on a stubborn mule.
Find
(a) the single force that is equivalent to the two forces shown
and
(b) the force a third person would have to exert on the mule to
make the net force equal to zero.
The forces are measured in units of newtons (N).
BRAINSTORMING-Definitions, concepts ,
principles and Discussion
(a) Gets the resultant vector R = F1 + F2 .
(b) Gets the equibrant vector E = -R .
Page 2 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 3 P20. The helicopter view in Figure P3.20 shows two people pulling on a
stubborn mule. Find
(a) the single force that is equivalent to the two forces shown and
(b) the force a third person would have to exert on the mule to make the net
force equal to zero.
The forces are measured in units of newtons (N).
y
Basic Solution
( a) The Resultant R = F1 + F2
F2=80N
In Get the Resultant picture, F2 and F1 are
added tip-to-tail to graphically get the
Resultant R. Then the component are drawn
to illustrate that
Rx = F1x +F2x
and
Ry = F1y + F2y .
F1x =
F2x =
F1y =
F2y =
o
F1cos1 = 120Ncos60
F2cos2 = -80Ncos75o
F1sin1 = 120Nsin60o
F2sin2 = 80Nsin75o
75o= 2
y
F1=120N
 1 = 60
Basic Vector Layout
F1 R
o

F2x F1x 1
x
= 120N* 0.50 = 60.0 N
= -80N*0.25882= -20.7 N
= 120N*0.866 = 103.9N
= 80N*0.96593=77.3 N
F1y
F2
R1y
F2y
 R=77.8
2
o
R1x
x
Get the Resultant
Rx = F1x +F2x = 60.0 – 20.7 = 39.3 N
Ry = F1y + F2y = 103.9 + 77.3 = 181.9 N
F1 R
R = sqrt (Rx2 + Ry2 ) = sqrt(39.32 + 181.92 ) = 185.4N
R vector = 185.4N /77.8o
R1y
F2y
 R=77.8
o
R1x
E (Eq
E = -R = -|R | /( R + 180o ) = 185.4 /77.8o +180o = 257.8o .
F2
t) = - R
(b) The Equilibrant, E, is shown the last (bottom) picture. It is equal in
magnitude and in the opposite direction as R.

F2x F1x 1
uibran
Tan R = Ry/Rx = 181.9N/39.3N = 4.61
R = tan-1 (4.61) = 77.8o
F1y
The Equibrant
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