# Static Equilibrium and downward force

• Meowzers
In summary, the conversation discussed solving a problem involving a boy carrying a sack on a stick balanced on his shoulder. The magnitude of the downward force the boy exerts with his hand was calculated using torque equations and the angle of 22 degrees from the horizontal. The correct calculation involved using cos 22 on both sides of the equation.
Meowzers
I just want to make sure that I approached the following problem correctly:

A boy carries a sack on one end of a very light stick that is balanced on his shoulder, at an angle of 22.0 degrees up from the horizontal. The mass of the sack is 7.00 kg, and it sits 1.20 m from his shoulder. If his hand is 0.350 m from his shoulder, on the other end of the stick, what is the magnitude of the downward force the boy exerts with his hand?

I drew somewhat of a force diagram, then:

Set x=force of hand
torque of hand + torque of wall = 0
rFcos(theta) + rFsin(theta) = 0
-(.350m)(x)(cos 22) + (1.2m)(9kg*9.8 m/s2)(sin 22) = 0
.3245x = 39.648
x = 122.18 N

Besides a couple of typos, why did you use the sin theta distance component as the moment arm for the downward force of the boy's hand? And oh if you use weight on right side of equation, you need to use it on left side also (i.e., multiply the mass by g.)

Last edited:
Well, in my force diagram, the 22 degree angle is right across from the downward force of the boy's hand. I guess they're just mirror images of each other...

So, the force of both the hand and the sack are Fsin(theta)?

torque of hand + torque of wall = 0
rFsin(theta) + rFsin(theta) = 0
-(.350m)(x)(sin 22) + (1.2m)(9kg*9.8 m/s2)(sin 22) = 0
.13111x = 39.648
x=302.4 N

Meowzers said:
Well, in my force diagram, the 22 degree angle is right across from the downward force of the boy's hand. I guess they're just mirror images of each other...

So, the force of both the hand and the sack are Fsin(theta)?

torque of hand + torque of wallyou mean sack? = 0
rFsin(theta) + rFsin(theta) = 0
-(.350m)(9.8)x(sin 22) + (1.2m)(9 or 7??kg*9.8 m/s2)(sin 22) = 0
.13111x = 39.648
x=302.4 N
Without a diagram, it is hard to understand where the 22 degree angle is; but in your statement you said it was up from the horizontal, in whichcase you should use cos 22 on both sides of the equation. There are 2 ways to calculate torques: The first is to take the magnitude of the force and multiply it by the perpendicular distance from the line of action of the force to the point you are taking moments about. In which case you have, for the sack torque,
$$F(r) cos\theta = 9(9.8)(1.2) cos 22$$.
OR, you can use the definition of torque
$$T =Frsin\theta$$, but here, $$\theta$$is the included angle between the force and displacemnt vectors, $$\theta = 68 degrees$$.. You will get the same answer; I prefer the first method.

Thanks for all your help :)

## 1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and there is no net force or torque acting on it.

## 2. How is static equilibrium different from dynamic equilibrium?

Static equilibrium occurs when an object is at rest, whereas dynamic equilibrium occurs when an object is in motion at a constant velocity.

## 3. What is the role of downward force in static equilibrium?

In static equilibrium, the downward force, also known as weight, is balanced by an equal and opposite upward force, such as a normal force or tension force.

## 4. What factors can affect static equilibrium?

The factors that can affect static equilibrium include the magnitude and direction of forces acting on the object, the object's shape and mass distribution, and the surface conditions on which the object is resting.

## 5. How is static equilibrium important in everyday life?

Static equilibrium is important in everyday life because it allows structures and objects to remain stable and balanced, preventing them from falling or collapsing. It is also crucial in the functioning of our bodies, as our muscles and bones rely on static equilibrium to maintain posture and perform movements.

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