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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Chapter1: Units, Trigonometry. And Vectors
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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
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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