Physics Torque Problem

[huang119]

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Example 3: A pole of length L is
connected to a hinge at point A on a
vertical wall, making an angle a with the
wall. A horizontal string connects to the
wall at point B and the end of the pble at
point C. A box of candy of mass m hangs
from a string at the end of the pole. (See
Figure 7- 12)

a What is the tension in the
horizontal string?
b. What is the magnitude of the
force of the wall on the pole at
point A?

I am having alot of difficulty understanding part B. If anyone could please explain part B to me that would be great. This example is from the Nova physics book in the chapter about torque. The solution is below:

 

[Quinoline]

There are 3 forces on point c. In the y direction, there are only 2 forces, Fs and mg. Since point c has no acceleration, the sum of the forces in the y direction equal 0. All of mg is in the y direction, but some of Fs is in the x direction, and some is in the y direction. The component in the y direction is found from using a right triangle with Fs as the hypotenuse, and Fsx and Fsy as the other 2 sides. The cos(a) = Fsy/Fs, so Fsy = cos(a)*Fs

Both Fsy and mg are in the -y direction, so summing these 2 forces to 0:
0=-Fs*cos(a)-mg
Then solve for Fs

 

[theonlytycrane]

Can the force of the wall on the pole be thought of as similar to a normal force from a surface to an object on the surface? I'm trying to better understand the forces present at points A and C.

 

[huang119]

Why do we choose point C as our starting point? I understand your explanation but I just don't know why we start at C or any other points as a matter of fact.

 

[Quinoline]

Can the force of the wall on the pole be thought of as similar to a normal force from a surface to an object on the surface? I'm trying to better understand the forces present at points A and C.

The wall exerts 2 forces on the pole, 1 is the normal force acting perpendicular to the wall, and the other is a friction force, acting up. These sum to a force in the opposite direction of Fs, of the same magnitude as Fs.

Why do we choose point C as our starting point? I understand your explanation but I just don't know why we start at C or any other points as a matter of fact.

I guess because at C you have enough information given to solve for Fs. Fs involves points A and C, and at point A you do not have enough information.

 

[theonlytycrane]

There are 3 forces on point c. In the y direction, there are only 2 forces, Fs and mg. Since point c has no acceleration, the sum of the forces in the y direction equal 0. All of mg is in the y direction, but some of Fs is in the x direction, and some is in the y direction. The component in the y direction is found from using a right triangle with Fs as the hypotenuse, and Fsx and Fsy as the other 2 sides. The cos(a) = Fsy/Fs, so Fsy = cos(a)*Fs

Both Fsy and mg are in the -y direction, so summing these 2 forces to 0:
0=-Fs*cos(a)-mg
Then solve for Fs

Is this in perspective of the string at point C? If so, I'm trying to figure out where the Fs force comes from. The string is in contact with the pole, so is this the Normal force from the pole onto the string?

Thanks for the help Quinoline!

 

[amy_k]

Yes, Fs is the force of the string on the pole. This is a bit like a crane arm problem. The force exerted on the pole by the string is the resultant force from T and mg.