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An athlete performs a leg extension on a machine using a $20-\mathrm{kg}$ mass at $A$ located $400 \mathrm{mm}$ away from the knee joint at center $O$ Biomechanical studies show that the patellar tendon inserts at $B$ which is $100 \mathrm{mm}$ below point $O$ and $20 \mathrm{mm}$ from the center line of the tibia (see figure). The mass of the lower leg and foot is $5 \mathrm{kg}$, the center of gravity of this segment is $300 \mathrm{mm}$ from the knee, and the radius of gyration about the knee is $350 \mathrm{mm}$. Knowing that the leg is moving at a constant angular velocity of 30 degrees per second when $\theta=60^{\circ}$ determine $(a)$ the force $\mathbf{F}$ in the patellar tendon, $(b)$ the magnitude of the joint force at the knee joint center $O$

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Physics 101 Mechanics

Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

Section 2

Constrained Plane Motion

Motion Along a Straight Line

Cornell University

Rutgers, The State University of New Jersey

Hope College

McMaster University

Lectures

04:34

In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

07:57

In mathematics, a position is a point in space. The concept is abstracted from physical space, in which a position is a location given by the coordinates of a point. In physics, the term is used to describe a family of quantities which describe the configuration of a physical system in a given state. The term is also used to describe the set of possible configurations of a system.

10:11

An athlete performs a leg …

05:10

During a concentric loadin…

02:22

During concentric oading o…

02:11

Problem-1 (10 Points): An …

03:57

A person is doing leg lift…

03:48

While exercising in a fitn…

03:22

The large quadriceps muscl…

07:05

01:15

04:14

01:11

01:22

If you stand on one foot w…

so here and here we have the leg rotating and lifting a mass that is situated on the essentially in the top part of, uh, the ankle. And so we have givens here. This is the mass of the of the mass of the of the weight that is that is being lifted by the leg. 20 kilograms multiplied by the mass of the lower leg and the foot five kilograms. Uh, we have the angular velocity about of rather for a the angular acceleration, and you have it in the free body diagram as well. But just for reference, these were the two angles fi right here and Seita right here. So we can first find the location of the center of mass. And so we can say that the X bar for this would be the center of mass equaling. Then there's some of the mess times the distance divided by the sum of the masses. This would be equaling them to m sub a multiplied by negative 0.4. This would be plus himself Al multiplied by negative points three multiplied by amsa a plus themselves out. And so this is going to be equal and then some from this gonna be equaling the negative 0.38 meters and we can say from point Oh, Torre words a queen a again we evaluate, we eat, we evaluate this at 0.0 and so now we are at the axis of rotation. We can say that then applying the kinetics the sum of forces in the X direction would be equaling then The mass times the linear acceleration in the X direction for the center of mass This is an Equalling negative 20 multiplied by 9.81 multiplied by co sign of 60 degrees minus five multiplied by 9.81 multiplied by co sign of 60 degrees plus o sub x plus UH, co sign of 20 degrees Equalling than M sub a GX or end multiplied by a sub GXE so we can find them the horizontal component of the reaction force equal and 122.63 minus f co sign of 20 degrees plus m. M, most played by a sub gxe. We do the exact same thing in the Y direction and we have the negative 20 multiplied by 9.81 multiplied by sine of 60 degrees minus five multiplied by 9.81 multiplied by sine of 60 degrees plus oh sub y plus F sign again of 20 degrees, equaling em. Multiply by, I said. Gee, why? And so we see that then the vertical component of the reaction forces equaling 212.39 minus f sign of 20 degrees plus and multiply by a sub g y. And so we can then see. But if we were to take the some of the moments some of the moments, not the force about the axis of rotation equaling, then the moment of inertia multiplied by the angular acceleration. We know that this is actually zero, and so we can say that then 20 multiplied by 9.81 20 kilograms, multiplied by 9.81 meters per second squared times sign of 60 degrees multiplied by 0.4. This would be the displacement between the point at which the forces, acting to the axis of rotation plus five five kilograms multiplied by 9.81 meters per second, squared again sign of 60 degrees multiplied by 600.3, minus af co sign of 20 degrees multiplied by point zero to minus F. Sign of 20 degrees multiplied by 0.1. And this is We know that this is equaling zero. And so all of these have been solved. All these have been substituted. Ah, and so weaken truly isolate f and we find that f the magnitude of the force f is equaling. Then 1500 and 22.9 Newtons moving in. Moving on. We have been more Kinnah Matics in the sense that we can say that the linear acceleration ah, for the center of mass equals the linear acceleration for the at the axis of rotation, which we know is zero plus the linear acceleration which we knows again. Zero rather the angular acceleration which we nose against zero cross product with the distance between the center of mass and the axis of rotation minus omega, the angular velocity for the axis of rotation squared, multiplied by the distance between the center of mass and the access of rotation. And so then we can see that then the X component of the linear acceleration would be 0.10 for two meters per second and we can actually solved this by just substituting in. If these two become zero, we simply have. Then we could have. This is equal to negative point 5 to 36 squared. I'm multiplied by negative 0.38 I had for the vector and so this would be I hat and we can say that then the linear acceleration in the Y direction would be zero meters per second squared. We can then substitute and for then the two reaction forces or to the to the components of the reaction force. So for the horizontal, a component of the reaction force, this would be equal to 122 0.63 minus 1522.9 multiplied by co sign of 20 degrees plus 25 multiplied by point 10 for two. So this is equaling negative 1305 0.85 Newtons. So this would be negative. In the diagram, we see that we drew it, um, to the right, essentially. So we know that this is to the left and then similarly, for the vertical component, we have 212.39 minus 1522.9. Sign of 20 degrees, plus 25 multiplied by zero. And so the vertical component is going to be equal to negative 300 an 8.47 Newtons. Of course this would be down. And when we say down again relative to our new Cartesian axes and then moving on, we simply have to use path agrarians and in order to find the magnitude so we can say that the magnitude of the reaction force at O would be equal to the square root. Oh said X squared, plus those of y squared, equaling square root of negative 1305.85 quantity squared plus negative 308 0.47 quantity squared and the fun. But then the magnitude of the reaction force at the hinge equals 1300 and 41.8 Newtons. So this would be our answer for part B that is Thean of the solution. Thank you for watching

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