Question

The device shown in the figure below (Figure 1) is one version of a Russell traction apparatus. It has two functions: to support the injured leg horizontally and at the same time provide a horizontal traction force on it. This can be done by adjusting the weight W and the angle 0. For this patient, his leg (including his foot) is 89.0 cm long (measured from the hip joint) and has a mass of 10.2 kg Its center of mass is 39.0 cm from the hip joint. A support strap is attached to the patient's ankle 13.0 cm from the bottom of his foot. What weight W is needed to support the leg horizontally? Express your answer in newtons. IVO AEQ ? W = N Figure < 1 of 1 Submit Request Answer v Part B Hip joint If the therapist specifies that the traction force must be 10.0 N horizontally, what must be the angle ? Express your answer in degrees. K 13.0 cm VALO w o =

Answer 1

To support the leg horizontally, a weight of 100.0 N is needed. For a horizontal traction force of 10.0 N, the angle must be 5.74°.

Step-by-step explanation:

To support the leg horizontally, we need to find the weight W. Since the leg has a mass of 10.2 kg, we can use the equation:

Weight = mass × gravitational acceleration

Weight = 10.2 kg × 9.8 m/s² = 100.0 N

Therefore, the weight W needed to support the leg horizontally is 100.0 N.

To calculate the angle θ for a horizontal traction force of 10.0 N, we can use the equation:

Traction force = weight × sin(θ)

10.0 N = 100.0 N × sin(θ)

Solving for θ:

θ = sin⁻¹(10.0 N / 100.0 N) = 5.74°

Therefore, the angle θ must be 5.74°.

Answer 2

To support the leg horizontally, a weight of 100.0 N is needed. For a horizontal traction force of 10.0 N, the angle must be 5.74°.

To support the leg horizontally, we need to find the weight W. Since the leg has a mass of 10.2 kg, we can use the equation:

Weight = mass × gravitational acceleration

Weight = 10.2 kg × 9.8 m/s² = 100.0 N

Therefore, the weight W needed to support the leg horizontally is 100.0 N.

To calculate the angle θ for a horizontal traction force of 10.0 N, we can use the equation:

Traction force = weight × sin(θ)

10.0 N = 100.0 N × sin(θ)

Solving for θ:

θ = sin⁻¹(10.0 N / 100.0 N) = 5.74°

Complete question :-

The device shown in the figure below is one version of a Russell traction apparatus. It has two functions: to support the injured leg horizontally and at the same time provide a horizontal traction force on it. This can be done by adjusting the weight and the angle For this patient, his leg (including his foot) is 95.0cm long (measured from the hip joint) and has a mass of 14.2kg . Its center of mass is 41.0cm from the hip joint. A support strap is attached to the patient's ankle 13.0 cm from the bottom of his foot.

Part A -
What weight is needed to support the leg horizontally?
W=___ N


Part B -
If the therapist specifies that the traction force must be 12.0 horizontally, what must be the angle
θ =_____


Part C -
What is the greatest traction force that this apparatus could supply to this patient's leg?
Fmax=_____ N

Part D
What is in that case?
θ=_____