Chapter 4 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 4 P20. The leg and cast in Figure P4.20 weigh
220 N (w1). Determine the weight w2 and the angle
 needed so that no force is exerted on the hip joint
by the leg plus the cast.
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 4 P20. The leg and cast in Figure P4.20 weigh 220 N (w1). Determine the weight w2
and the angle  needed so that no force is exerted on the hip joint by the leg plus the cast.
BRAINSTORMING-Definitions, concepts , principles and Discussion
1. In the picture, use
Newton’s 3rd Law of
Motion to replace the
physical connections from
pulleys with actionreaction pairs as
illustrated.
-T2
-T1
T1
T2
2. Then sketch the FBD
(Free Body Diagram) of
the joint, as shown.
3. Set up the vector
diagram by drawing the x
and y axes and sketching
T1 and T3 accurately to
scale.
4. Complete the
Graphical Solution for T2.
5. Use the Component
Method to solve for T2
analytically. It is a bit
challenging. Use 4,
above, as a help.
(Hint: T1x = T1cos1)
y
T2
T1
T1=110N
1 =
T2=w2
T1
2 =
x
T3
T1
T3=220N
T3=220N
FBD of
Joint
Vector
Diagram
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Graphical
Solution
for T2
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 4 P20. The leg and cast in Figure P4.20 weigh 220 N
(w1). Determine the weight w2 and the angle  needed so
that no force is exerted on the hip joint by the leg plus the
cast.
Basic Solution
(Definitions, concepts , principles and Discussion
included)
Given: T3 = w1 = 220N , T1 = 110 N, 1 = 40o .
Find:
2 =  , T2 = w2 .
Newton’s 3rd Law of Motion has already been used to isolate the joint from
its physical connections by way of action-reaction pair.
1
Use components. The x axis is the horizontal and the y
axis the vertical directions.
2=
T1x= T1cos T 2x= T 2cos
joint
T3=W1=220N
Equillibrium
So ,F=0
The vector
T2
sum is a
closed triangle.
The magnitude
and direction of
T2 can be scaled
off the diagram
Newton’s 2nd Law of Motion
T2
T 2y= T2sin
T1
T1y= T 1sin
y
x
+->Fx = 0. –T1x + T2x = -T1cos1 + T2cos2 = 0. (1)
Up+->Fy = 0. T1y + T2y – W1 = 0
T1sin1 + T2sin2 – W1 = 0. (2)
T2 and 2 are both unknown. They can be eliminated.
These two variables occur together as T2sin2 and
T2cos2 . This suggests solving for the angle 2 by
Solving for
tan  2 
T1
T2 sin θ 2
W  T sin θ1
 1 1
,
T2 cos θ 2
T1 cos θ1
the numerator(top) follows from (2)
T3
T1
Graphical Solution
for T2Completed
and the denominator(bottom) from (1)
Then the angle follows from the arctan ;
θ 2 = α = arctan (
=arctan(
= arctan(
W1  T1 sin θ1
)
T1 cos θ1
220N -110Nsin40o
)
110N cos40o
149.3
) = arctan(1.772)=60.6o
84.3
Now to get T2 , Equation (1) gives
cosθ1
cos40o
0.766
T2 = T1
= 110N*
=110N*
=172 N.
o
cosθ 2
0.491
cos60.6
Within the significant figures, 2 =  = 61o , and T2 = w2 = 1.7 x 102 N.
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