The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the weight of the forearm and the book. The triceps are assumed to be relaxed. The lower panel, is an approximately equivalent mechanical system with the pivot at the elbow joint. There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is FB; that of the elbow joint is FE; that of the weights of the forearm is wa, and its load is wb. Two of these are unknown( FB and FE). Solve for FB after answering the questions below. Triceps muscle FE 4.0 cm 16 cm Fa 1₂ Biceps muscle m₂ = 2.5 kg CG W₁ 38 cm CG Wa 13 (a) (b) PHYSICS m₂ = 4.0 kg CG W₂ CG W₂ Free-body diagram Wat FB W₂

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  1. elbow
  2. tip of the finger 
  3. any position different from the elbow on the forearm
The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the
weight of the forearm and the book. The triceps are assumed to be relaxed. The lower panel, is an
approximately equivalent mechanical system with the pivot at the elbow joint.
There are four forces acting on the forearm and its load (the system of interest). The magnitude of
the force of the biceps is FB; that of the elbow joint is FE; that of the weights of the forearm is wa, and
its load is Wb. Two of these are unknown( FB and FÉ). Solve for FB after answering the questions
below.
Triceps
muscle
FE
4.0
cm
1₁
16 cm
FB
12
Biceps
muscle
m₂ = 2.5 kg
CG
Wa
38 cm
CG
Wa
13
(a)
(b)
PHYSICS
m₂ = 4.0 kg
CG
W₂
CG
W₂
Free-body
diagram
Wa
FB
Wb
Transcribed Image Text:The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the weight of the forearm and the book. The triceps are assumed to be relaxed. The lower panel, is an approximately equivalent mechanical system with the pivot at the elbow joint. There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is FB; that of the elbow joint is FE; that of the weights of the forearm is wa, and its load is Wb. Two of these are unknown( FB and FÉ). Solve for FB after answering the questions below. Triceps muscle FE 4.0 cm 1₁ 16 cm FB 12 Biceps muscle m₂ = 2.5 kg CG Wa 38 cm CG Wa 13 (a) (b) PHYSICS m₂ = 4.0 kg CG W₂ CG W₂ Free-body diagram Wa FB Wb
(a) From figure (b), pick the first condition of equilibrium from below:
A. FB +FE+ Wa - Wb = 0
B. FB - FE - W₂ - Wb = 0
C. FB-FE-Wa+Wb = 0
D. FB-FE+Wa+Wb = 0
First condition of equilibrium is
(b) What point should you calculate the torque in order to get rid of the second unknown FE.
(c) From figure (b), pick the second condition of equilibrium from below:
A. r₁ FB +r₁ FE + r₂ Wa - 13 Wb = 0
B. ₁ FB + r₂ Wa +r3 Wb = 0
C. r₁FB - r₂ Wa + r3 W₁ = 0
D. r₁ FB
2 War3 Wb=0
Second condition of equilibrium is
-
(d) Solve for FB.
Enter to 2 significant figures
FB =
N
(e) Compare the FB to the combined weights it supports.
Enter to 2 significant figures
The force in the biceps is
FB
Wa+Wb
=
times larger than the weights it supports.
Transcribed Image Text:(a) From figure (b), pick the first condition of equilibrium from below: A. FB +FE+ Wa - Wb = 0 B. FB - FE - W₂ - Wb = 0 C. FB-FE-Wa+Wb = 0 D. FB-FE+Wa+Wb = 0 First condition of equilibrium is (b) What point should you calculate the torque in order to get rid of the second unknown FE. (c) From figure (b), pick the second condition of equilibrium from below: A. r₁ FB +r₁ FE + r₂ Wa - 13 Wb = 0 B. ₁ FB + r₂ Wa +r3 Wb = 0 C. r₁FB - r₂ Wa + r3 W₁ = 0 D. r₁ FB 2 War3 Wb=0 Second condition of equilibrium is - (d) Solve for FB. Enter to 2 significant figures FB = N (e) Compare the FB to the combined weights it supports. Enter to 2 significant figures The force in the biceps is FB Wa+Wb = times larger than the weights it supports.
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