Question

Consider the flagpole in the figure below. If the flagpole has a mass of 18 kg and length 10 m and the angle the cable makes with the pole is ϕ = 22°, what are the magnitude and direction of the force exerted by the hinge (at point P) on the flagpole? Assume the mass of the pole is distributed uniformly. magnitude N direction ° counterclockwise from the +x-axis

Answer 1

Assuming the image to be the below figure::----

For these kinds of questions, you may have to analyze the forces, or the torques, or both.

Let's start with the forces and see where we get.

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The flagpole is not accelerating, which means the net force on it is zero. That fact allows us to write some equations about the force. It's easiest to consider the horizontal and vertical forces separately.

Let:

T = tension in cable;
Hx = horizontal component of force exerted by hinge;
Hy = vertical component of force exerted by hinge

Horizontal forces:
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1. Horizontal component of tension (-Tcosφ) (negative because it points to the left)
2. Horizontal component of hinge force (Hx)

The net horizontal force is zero, so:

-Tcosφ + Hx = 0 or   Hx = Tcosφ

Vertical forces:
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1. Vertical component of tension (Tsinφ)
2. Vertical component of hinge force (Hy);
3. Weight of flagpole (-mg) (negative because it points downward)

Net vertical force is zero, so:

Tsinφ + Hy + (-mg) = 0 or Hy = mg - Tsinφ

A walkthrough to all steps calculated above:::: -------
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Hx = Tcosφ
Hy = mg - Tsinφ

But those are 2 equations in 3 unknowns (Hx, Hy and T); so you can't find a solution yet. This means we have to look at torques too.

You can consider the torques about any axis you choose; but it's convenient to choose the pin. That way, the hinge force "H" doesn't contribute to the torque (because it acts through the axis); making the calculation part much easier .

Clockwise torque:
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Due to weight of flagpole; assumed to act at its center. So:
clockwise torque = (mg)(L/2) (where "L" = length of pole, 10m)

Counter-clockwise torque:
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Due to tension of cable:
counter-clockwise torque = (T)(L)sinφ

Since the pole has no rotational acceleration, the two torques cancel (they have the same magnitude); so:

(mg)(L/2) = (T)(L)sinφ

Solve for "T":
T = mg/(2sinφ)

A walkthrought to above analysis::::::---------:
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Hx = Tcosφ
Hy = mg - Tsinφ
T = mg/(2sinφ)

These are 3 equations in 3 unknowns (Hx, Hy and T). SO it seems now we can use the ordinary algebraic equation solving tchniques to solve for all 3 (but according tothe question you only have to take care about Hx and Hy).

Once you know Hx and Hy, you can figure out the magnitude and direction of "H":

Magnitude of H is given by: H² = (Hx)² + (Hy)²
Direction θ of H is given by: tanθ = Hy/Hx.

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Sum of the moments about the hinge = 0

0 = T*sin(22)*10 - m*g*5 (mg5 because the weight will be at the center of the flagpole)

T = m*g/(2*sin(22))

Now just substitute the values given in the question in my solution and you reach your solution.

NOTE:-PLEASE GIVE DIAGRAM NEXT TIME