Find the force a distance from a fixed point

In summary: Since T and R are perpendicular to the rod, they cannot both act on the same side of the rod. T would act at the point of rotation and R would act 30.29cm away.
  • #1
ferociablejbear
11
0

Homework Statement


A person is standing on tiptoe, with the total weight supported by the force on the toe. A mechanical model for the situation is shown, where T is the force in the Achilles tendon and R is the force on the foot due to the tibia. Find the value of T. Assume the total weight is 484N. Answer in units of N.

R = unknown
T = unknown
Distance of T from point of rotation = 32cm = l2
Distance of R from point of rotation = 30.29cm = l1
Angle of T with respect to force acting at point of rotation = 21.2 degrees = θ
Angle of R with respect to force acting of point of rotation = 14.4 degrees = φ

It's a statics problem, so there's no rotational or translational motion. Also, the forces are orthogonal to the radial axis.

Homework Equations


ΣFy=Tcosθ-Rcosφ+484N=0
Στ=T*l2-Rl1=0

The Attempt at a Solution


Since there are only two unknowns, I took the relationship between the two torques to find R in terms of T, R=T(l2/l1). I then plugged this into the equation for the sum of forces in the y-direction, and I got the answer 672.5992756N for T, but this apparently isn't correct. I really can't figure out what I'm doing wrong, so any help would be appreciated.
 
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  • #2
ferociablejbear said:

Homework Statement


A person is standing on tiptoe, with the total weight supported by the force on the toe. A mechanical model for the situation is shown, where T is the force in the Achilles tendon and R is the force on the foot due to the tibia. Find the value of T. Assume the total weight is 484N. Answer in units of N.

R = unknown
T = unknown
Distance of T from point of rotation = 32cm = l2
Distance of R from point of rotation = 30.29cm = l1
Angle of T with respect to force acting at point of rotation = 21.2 degrees = θ
Angle of R with respect to force acting of point of rotation = 14.4 degrees = φ

It's a statics problem, so there's no rotational or translational motion. Also, the forces are orthogonal to the radial axis.

Homework Equations


ΣFy=Tcosθ-Rcosφ+484N=0
Στ=T*l2-Rl1=0

The Attempt at a Solution


Since there are only two unknowns, I took the relationship between the two torques to find R in terms of T, R=T(l2/l1). I then plugged this into the equation for the sum of forces in the y-direction, and I got the answer 672.5992756N for T, but this apparently isn't correct. I really can't figure out what I'm doing wrong, so any help would be appreciated.
The question refernces a diagram which you have not included. If you cannot upload a diagram, at least try to describe it in detail.
 
  • #3
It's a diagram of a rod with normal force equal to the weight acting at the point of rotation. The two forces T and R are acting perpendicularly to the rod at given distances in opposite directions. Assuming the normal force defines the positive y-axis, force T and force R make the quoted angles with the y-axis.
 
  • #4
ferociablejbear said:
It's a diagram of a rod with normal force equal to the weight acting at the point of rotation. The two forces T and R are acting perpendicularly to the rod at given distances in opposite directions. Assuming the normal force defines the positive y-axis, force T and force R make the quoted angles with the y-axis.
I can't make sense of that. If T and R are perpendicular to the rod then all three forces are parallel, and all perpendicular to the rod so where do these angles come in?
Where is the point of rotation of the rod?
 
  • #5
The angles are to allow resolution into components of force along the x and y axis. The normal force acts wholly in the y-direction and is not perpendicular to the rod. The axis of rotation is at the point where the normal force is acting. It is 30.29cm from R and 32cm from T. The normal force cannot participate in torque, but the rod isn't fixed, so forces in the x and y directions need to balance.
 
  • #6
ferociablejbear said:
The angles are to allow resolution into components of force along the x and y axis. The normal force acts wholly in the y-direction and is not perpendicular to the rod. The axis of rotation is at the point where the normal force is acting. It is 30.29cm from R and 32cm from T. The normal force cannot participate in torque, but the rod isn't fixed, so forces in the x and y directions need to balance.
Then you should not have called it a normal force. A normal force, by definition, acts at right angles to the surface or line that it acts on. I guess you meant a vertical force.
Assuming that is vertically down, which of R and T, if either, act on the same side of the rod as that?
 
  • #7
I just noticed that in your original post you specified angles to the rod for R and T, but in post #3 that they are perpendicular to the rod. I assume the OP is the correct version.
 
  • #8
They are acting in opposite directions with no rotation. The rod is in contact with a surface, hence normal force. That's the language of the question, not my own. If you mean which is acting in the same direction as the normal force, then the answers is neither. In the case of y components, R is acting in the opposite direction of the normal force and T is acting in the same direction as the normal force.
 
  • #9
The angle is with respect to the y axis. They are perpendicular to the rod, which is at some unknown angle to the x and y axes.
 
  • #10
ferociablejbear said:
The angle is with respect to the y axis. They are perpendicular to the rod, which is at some unknown angle to the x and y axes.
Is this a straight rod? If so, and they are both perpendicular to it, then they must be parallel, but in the OP you specified angles of 21.2 and 14.4 to the same reference direction, so they cannot be parallel.
 
  • #11
They are antiparallel.
 
  • #12
cbf0674f-2ff0-4777-8d1c-44cfd6709ec8.gif
 
  • #13
That's analagous but with different numbers, obviously.
 
  • #14
ferociablejbear said:
They are antiparallel.
They cannot be antiparallel either. The angle between them must be 21.2-14.4, or 180 minus that etc.
 
  • #15
So, I was able to get the right answer by simply using the sum of the forces in the x and y direction rather than using torque, but I'd still like to understand how to solve it this way. What would the relationship of the torques be? They are obviously equal, and I get that you're saying some component would need to be parallel to the rod, but I don't understand how to discover that relationship. Would it be Tl2-Rcosφl1=0? It seems like it would need to be that, but I don't get the same answer as when it's just simple force balancing.
 
  • #16
ferociablejbear said:
So, I was able to get the right answer by simply using the sum of the forces in the x and y direction rather than using torque, but I'd still like to understand how to solve it this way. What would the relationship of the torques be? They are obviously equal, and I get that you're saying some component would need to be parallel to the rod, but I don't understand how to discover that relationship. Would it be Tl2-Rcosφl1=0? It seems like it would need to be that, but I don't get the same answer as when it's just simple force balancing.
Well, I'm still unclear on the exact diagram for the question that kicked off this thread, so let's discuss it in relation to the diagram in post #12. Despite the appearance of R and T there, I presume they are not parallel, nor antiparallel. Rather, T makes an angle theta to the vertical, the rod makes angle theta to the horizontal, and R makes an angle 15 degrees to the vertical.
So there are three unknowns, the magnitudes of R and T and the angle theta.

For a 2D static arrangement there are three equations available for each rigid body. Usually people take two linear and and one moments equation, but other combinations can be used as long as they are algebraically independent. In particular, three linear equations won't do it, since any two implies the third.

From what I have been able to glean about the set up in post #1, you know the angles of R and T to the vertical but perhaps not the angle of the rod. If you do know the angle of the rod then the question is overspecified, perhaps even self-contradictory. If you do not know it then you must either use three equations or find two which do not involve that angle. That last is what I think you have done. Since you know the angles between all the forces, you can get two linear equations that do not involve the angle of the rod. If you want to find the angle of the rod then you will need to take moments about somewhere.
 
  • #17
Using the data in the diagram in post #12 where l1=0.18m, l2=0.25m, φ=15°, and assuming n is known, would a valid system of equations to determine T be:
Στ: Tl2-Rcosφl1=0
∑Fy: Tcosθ+n-Rcosφ=0
∑Fx: Tsinθ=Rsinφ?

Or would the angle for the component of force R orthogonal to the rod need to be cos(|φ-θ|)?
 
  • #18
ferociablejbear said:
Using the data in the diagram in post #12 where l1=0.18m, l2=0.25m, φ=15°,
No, as I posted, theta in that diagram is unknown. Although R and T look to be antiparallel, they are not. There is enough information to deduce theta, and it will not be 15 degrees. This is easy to prove. By moments about the left end of the rod, R and T cannot be equal in magnitude. From the horizontal balance of forces, they must have the same horizontal component. So they must make different angles with the vertical.
 

Related to Find the force a distance from a fixed point

1. What is the equation for finding the force at a distance from a fixed point?

The equation for finding the force at a distance from a fixed point is F = (G * m1 * m2) / r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.

2. What is the significance of the distance in calculating the force?

The distance between two objects is a crucial factor in calculating the force between them. As the distance increases, the force decreases, and vice versa. This is because the force of gravity follows an inverse square law, meaning that the force is inversely proportional to the square of the distance between the objects.

3. How does the mass of the objects affect the force at a distance?

The mass of the objects has a direct impact on the force at a distance. The greater the mass of the objects, the stronger the force between them will be. This is because the force of gravity is directly proportional to the masses of the objects involved.

4. Can the distance from a fixed point affect the direction of the force?

No, the distance from a fixed point does not affect the direction of the force. The force of gravity always acts in a straight line between the two objects, regardless of the distance between them.

5. How can the force at a distance be calculated experimentally?

To calculate the force at a distance between two objects experimentally, you would need to measure the masses of the objects, the distance between them, and the gravitational constant. Then, you can plug these values into the equation F = (G * m1 * m2) / r^2 to determine the force between the objects.

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