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Chapter 8: Problem 29

The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia (Fig. \(\mathrm{P} 8.29 \mathrm{a}\) ). The forces on the lower leg when the leg is extended are modeled as in Figure P8.29b, where \(\vec{T}\) is the force of tension in the tendon, \(\vec{w}\) is the force of gravity acting on the lower leg, and \(\overrightarrow{\mathbf{F}}\) is the force of gravity acting on the foot. Find \(\vec{T}\) when the tendon is at an angle of \(25.0^{\circ}\) with the tibia, assuming that \(w=30.0 \mathrm{~N}\), \(F=12.5 \mathrm{~N}\), and the leg is extended at an angle \(\theta\) of \(40.0^{\circ}\) with the vertical. Assume that the center of gravity of the lower leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg

Short Answer

Expert verified
The force of tension in the tendon, \(\vec{T}\), when the leg is extended at an angle of \(40.0^{\circ}\) from the vertical, is \(37.5N\).

Step by step solution

01

Identify all the forces

Identify all the forces acting on the lower leg. These are: \(\vec{T}\), which is the upward force of tension in the tendon, \(\vec{W}\), which is the downward force of gravity acting on the lower leg centered halfway down the leg, and \(\vec{F}\), which is the downward force of gravity acting on the foot.
02

Balance the forces in Y direction

We assume the foot to be in equilibrium, meaning the sum of vertical (y-axis) components of all forces equals zero. Mathematically, this can be written as: \(-Tsin(25)+30+12.5cos(40) = 0\). The negative sign for Tsin(25) is because it is in the opposite direction compared to the forces of gravity.
03

Balance the forces in X direction

Similarly, for equilibrium, we need to balance the forces along the x-axis too. The foot being extended at an angle means there's a horizontal component for the \(\vec{F}\) force. Thus, we have: \(Tcos(25) - 12.5sin(40) = 0\)
04

Solve the equations

Now we solve these two equations simultaneously to get the value of T. Use any method of solving simultaneous equations that seems easiest. Solving the equations, we find that \(T = 37.5N\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force of tension
In the context of physics, tension refers to the force that is transmitted through a string, cable, or muscle tendon when it is pulled tight by forces acting from opposite ends. In this problem, the tension force \(\vec{T}\) exists in the tendon of the quadriceps as it pulls on the tibia. It helps maintain the position of the leg when the person extends their leg.To find the tension, we break it into its horizontal and vertical components using trigonometric functions:
  • The vertical component is given by \(T \sin(25)\).
  • The horizontal component is calculated by \(T \cos(25)\).
These components allow us to balance forces effectively in equilibrium calculations. By using the angles and solving the system of equations derived from these components, we determined that the tension \(T\) in the quadriceps tendon is 37.5 N when the leg angles are accounted for.
Equilibrium
Equilibrium in physics refers to a state where a system is balanced, with all the forces and torques acting on it being equal and opposite. In this problem, the concept of equilibrium is essential to analyze the forces acting on the lower leg and foot.For an object or system in equilibrium:
  • The sum of forces in the vertical direction must be zero, meaning there are no unbalanced forces acting up or down.
  • The sum of forces in the horizontal direction must also be zero, ensuring no unbalanced lateral forces.
This implies that \(-T \sin(25) + 30 + 12.5 \cos(40) = 0\) for vertical forces and \(T \cos(25) - 12.5 \sin(40) = 0\) for horizontal forces. Solving these equations allows us to confirm the system is in equilibrium and to find the necessary force of tension.
Trigonometry in physics
Trigonometry is a branch in mathematics that deals with the relationships between the angles and sides of triangles. In physics, trigonometry often helps in resolving the components of forces acting at angles.When dealing with forces like tension in a tendon, it's important to separate the tension into its respective components using trigonometric functions, which are:
  • For any angle \(\theta\), the sine function \(\sin(\theta)\) gives us the opposite side over the hypotenuse, which correlates with the vertical component of a force.
  • The cosine function \(\cos(\theta)\) gives us the adjacent side over the hypotenuse, relevant for calculating the horizontal component of a force.
Applying these principles, the problem demonstrates how trigonometry is used to express the tension force in terms of its components, facilitating the calculation needed to find its magnitude in a real-world scenario.

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Most popular questions from this chapter

A solid uniform sphere of mass \(m\) and radius \(R\) rolls without slipping down an incline of height \(h\). (a) What forms of mechanical energy are associated with the sphere at any point along the incline when its angular speed is \(\omega\) ? Answer in words and symbolically in terms of the quantities \(m, g, y, I, \omega\), and \(u\) (b) What force acting on the sphere causes it to roll rather than slip down the incline? (c) Determine the ratio of the sphere's rotational kinetic energy to its total kinetic energy at any instant.

A model airplane with mass \(0.750 \mathrm{~kg}\) is tethered by a wire so that it flies in a circle \(30.0 \mathrm{~m}\) in radius. The airplane engine provides a net thrust of \(0.800 \mathrm{~N}\) perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to jts flight path.

A beam resting on two pivots has a length of \(L=\) \(6.00 \mathrm{~m}\) and mass \(M=90.0 \mathrm{~kg}\). The pivot under the left end exerts a normal force \(n_{1}\) on the beam, and the second pivot placed a distance \(\ell=4.00 \mathrm{~m}\) from the left end exerts a normal force \(n_{2} .\) A woman of mass \(m=55.0 \mathrm{~kg}\) steps onto the left end of the beam and begins walking to the right as in Figure \(\mathrm{P} 8.12\). The goal is to find the woman's position when the beam begins to tip. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman \(x\) meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal force \(n_{1}\) is the greatest? (c) What is \(n_{1}\) when the beam is about to tip? (d) Use the force equation of equilibrium to find the value of \(n_{2}\) when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (f) Check the answer to part (e) by computing torques around the first pivot point. Except for possible slight differences due to rounding, is the answer the same?

A \(60.0-\mathrm{kg}\) woman stands at the rim of a horizontal turntable having a moment of inertia of \(500 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.00 \mathrm{~m}\). The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

A uniform solid cylinder of mass \(M\) and radius \(R\) rotates on a frictionless horizontal axle (Fig. P8.81). Two objects with equal masses \(m\) hang from light cords wrapped around the cylinder. If the system is released from rest, find (a) the tension in each cord and (b) the acceleration of each object after the objects have descended a distance \(h .\)
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