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92.BIO In an emergency situation,
a person with a broken forearm
ties a strap from his hand to clip
on his shoulder as in Figure P8.92.
His 1.60-kg forearm remains in a
horizontal position and the strap
makes an angle of 0 = 50.0° with
the horizontal. Assume the fore-
arm is uniform, has a length of
l = 0.320 m, assume the biceps
muscle is relaxed, and ignore the
mass and length of the hand. Find
(a) the tension in the strap and
(b) the components of the reac-
tion force exerted by the humerus
on the forearm.

92BIO In an emergency situation a person with a broken forearm ties a strap from his hand to clip on his shoulder as in Figure P892 His 160kg forearm remains in class=

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mickymike92

Based on the principle of equilibrium of forces:

  • the tension T, in the strap is 10.2 N
  • thehorizontal component of the reaction, Rx is 6.56 N
  • thevertical component of the reactionary force, Ry is 7.84 N

What is tension?

Tension is a type of force that is transmitted through a flexible medium such as a cable, wire or string pulled tight by forces acting from opposite ends.

The tension in the strap is calculated as follows:

Taking sum of moments from the right end as equal to zero:

mg × l/2 = Tsinθ × l

T = mg/2sinθ

T = 1.60 × 9.8 / 2 sin 50°

T = 10.2 N

Calculating the components of the reac-

reac-tionary force:

  • Horizontal component is Rx
  • Vertical component is Ry

From Newton's third law:

Rx = Tcosθ

Rx = 10.2 × cos 50°

Rx = 6.56 N

Ry - Tsinθ = 0

Ry = Tsinθ

where;

T = mg/2sinθ

Ry = mgsinθ/2sinθ

Ry = mg/2

Ry = 7.84 N

Learn more about tension at: https://brainly.com/question/24994188

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