
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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I need help understading this and the calculations step by step
the answer is in blue
and i need help how the got 43 n step by step please

Transcribed Image Text:When you bend your knee, the quadriceps muscle is
stretched. This increases the tension in the quadriceps ten-
don attached to your kneecap (patella), which in turn in-
creases the tension in the patella tendon that attaches your
kneepcap to your lower leg bone (tibia). Simultaneously,
the end of your upper leg bone (femur) pushes outward on
the patella. In the orientation shown in the figure, the mag-
Quadriceps
Tendon
Femur
Patella
nitude of the tension in each tendon is 60 N. What is the
magnitude of the force being exerted on the patella by the
femur in this situation? Assume that the two tendon forces
Tibia
42°
shown and the force exerted by the femur are the only three
forces. (Hint: this image was taken with the person at rest.)
Patella Tendon
+ 60
Here's a free body diagram showing the two tendon forces, with
the lower one split into components. Since the image is taken
at rest, we need zero net force both vertically and horizontally.
That means we need a horizontal component of 60 sin(42) to
the right, and a vertical component of 60 – 60 cos(42) pointing
down. Calculate those two values, and then use the Pythagorean
theorem to combine them as the magnitude of a single vector.
60 sin(42)
' 60 cos(42)
The answer is | 43 N
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