Download Solutions to Selected Introductory Problems: Week 2

Solutions to Selected Introductory and Additional Problems: Week 14
SOLUTIONS TO SELECTED INTRODUCTORY PROBLEMS (Ch 13)
3. A 23 kg boy sits 1.5 m from the axis of rotation of a seesaw. At what distance from the axis
of rotation must a 21 kg boy be positioned on the other side of the axis to balance the seesaw?
Fd⊥1 = Fd⊥2
1.5(23) = 21d⊥2
d⊥2 = 1.64 m
4. How much force must be produced by the biceps brachii at a perpendicular distance of 3 cm
from the axis of rotation at the elbow to support a weight of 200 N at a perpendicular distance of
25 cm from the elbow?
∑T = 0
Fd⊥1 = Fd⊥2
F(3) = 200(25)
F = 1667 N
5. Two people push on opposite sides of a swinging door. If A exerts a force of 40 N at a
perpendicular distance of 20 cm from the hinge and B exerts a force of 30 N at a perpendicular
distance of 25 cm from the hinge, what is the resultant torque acting at the hinge, and which way
will the door swing?
T = Fd⊥
Ta = 40 N(0.2 m) = 8 Nm
Tb = 30 N(0.25 m) = 7.5 Nm
8 – 7.5 = 0.5 Nm in A direction
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9. A 10 kg block sits motionless on a table in spite of an applied horizontal force of 2 N. What
are the magnitudes of the reaction force and friction force acting on the block?
R = 10(9.81) = 98.1 N
Fs = 2N
equal to horizontal force because block is motionless
SOLUTIONS TO SELECTED ADDITIONAL PROBLEMS (Ch 13)
4. A hand exerts a force of 90 N on a scale at 32 cm from the joint center at the elbow. If the
triceps attach to the ulna at a 90° angle and at a distance of 3 cm from the elbow joint center, and
if the weight of the forearm and hand is 40 N with the hand/forearm CG located 17 cm from the
elbow joint center, how much force is being exerted by the triceps? (733.3 N)
∑Me = 0
0 = (90 N)(32 cm) - Fm(3 cm) - (40 N)(17 cm)
Fm = 733.3 N
5. A patient rehabilitating a knee injury performs knee extension exercises wearing a
15 N weight boot. Calculate the amount of torque generated at the knee by the weight boot for
the four positions shown, given a distance of 0.4 m between the weight boot's CG and the joint
center at the knee. (a. 0, b. 3 N-m, c. 5.2 N-m, d. 6 N-m)
(a)
Tk = (15 N)(0.4 m)(sin 0°) = 0
(b)
Tk = (15 N)(0.4 m)(sin 30°) = 3 N-m
(c)
Tk = (15 N)(0.4 m)(sin 60°) = 5.2 N-m
(d)
Tk = (15 N)(0.4 m)(sin 90°) = 6 N-m
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7. A worker leans over and picks up a 90 N box at a distance of 0.7 m from the axis of rotation in
her spine. Neglecting the effect of body weight, how much added force is required of the low
back muscles with an average moment arm of 6 cm to stabilize the box in the position shown in
the text? (1050 N)
∑Me = 0 = (90 N)(0.7 m) - Fm(0.06 m)
Fm = 1050 N
9. A therapist applies a lateral force of 80 N to the forearm at a distance of 25 cm from the axis
of rotation at the elbow. The biceps attaches to the radius at a 90° angle and at a distance of 3 cm
from the elbow joint center. (a) How much force is required of the biceps to stabilize the arm in
this position? (b) What is the magnitude of the reaction force exerted by the humerus on the
ulna? (a. 666.7 N, b. 586.7 N)
(a)
∑Me = 0 = Fm(3 cm) - (80 N)(25 cm)
Fm = 666.7 N
(b)
∑Me = 0 = 80 N - 666.7 N + Fr
Fr = 586.7 N
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