# Can Static Equilibrium Be Maintained Without Friction in This Beam Scenario?

• Corey
In summary: What about the static friction between the ground and the pole, isn't that another horizontal force?No, the static friction is only between the beam and the ground.
Corey
A uniform vertical beam of mass 40 kg is acted on by a horizontal force of 520 N at its top and is held, in the vertical position, by a cable as shown.

a) Draw a free-body diagram for the beam, clearly labeling all of the forces acting on it.

b)Calculate the tension in the cable?

c)Determine the reaction forces acting on the beam by the ground?

Sorry I could not post the picture but its a vertical beam that is 5 m long and the force is coming out at the horizontal at the top of the beam. The cable is connected 3 m up the beam with a theta of 28 degrees.

Formulas:

Fx=0
Fy=0
Torque=0

So what I have done is broken the components of tension into its x and y components. Horizontally it is Tcos theta and vertically it is Tsin tetha

so:

Fy=0
Fn-mg-Tsin theta= 0
Tsin theta=0 (because Fn and mg as equal and opposite)

Fx=0
F- Tcos theta=0
F= Tcos theta
T= F / cos 28
T= 588.93

is this correct for the part b? Not using the torque equation is kind of throwing me off because of how it is connected. Any suggestions on if I did it right or not?

Corey said:
The cable is connected 3 m up the beam with a theta of 28 degrees.

Do you mean 28 degrees above the horizontal? (An angle is meaningless unless related to something)

Corey said:
Fx=0
F- Tcos theta=0
F= Tcos theta
T= F / cos 28
T= 588.93is this correct for the part b? Not using the torque equation is kind of throwing me off because of how it is connected. Any suggestions on if I did it right or not?

There is torque involved. You should have put Torque = 0 (instead of: horizontal force = 0)

EDIT:
The horizontal force = 0 is for part C

Torque = 0 is what applies to part B

Sorry about that, yes it is 28 degrees above the horizontal

I just thought it would be unnecessary because we could of isolated T already but:

T=0
choosing the bottom portion of the beam as the pivot point

T=0
F(x) - Tcos Theta (x)=0
Fx= Tcos theta (x)
540N (5m) = T cos (28) (3m)
T=1019

Corey said:
Fy=0
Fn-mg-Tsin theta= 0
Tsin theta=0 (because Fn and mg as equal and opposite)

If $Tsin(θ) = 0$ then $T=0$ which isn't true.

The flaw in your reasoning is that $F_n$ and $mg$ are not equal and opposite. $F_n$ is less than $mg$ because some of the weight is "taken off" by $Tsin(θ)$
(Imagine holding something so that it just barely touches the ground. It will be touching the ground, but it's weight will be supported by your hand, so $F_n$ would be zero. Same thing applies here, except $Tsin(θ)$ only supports some of the weight, not all of it.)

Corey said:
I just thought it would be unnecessary because we could of isolated T already but:

T=0
choosing the bottom portion of the beam as the pivot point

T=0
F(x) - Tcos Theta (x)=0
Fx= Tcos theta (x)
540N (5m) = T cos (28) (3m)
T=1019
How can T have such different answers?

Is T torque, or is it tension ?

dont you have to use the torque equation to isolate the tension aspect and then solve?

Corey said:
dont you have to use the torque equation to isolate the tension aspect and then solve?

What SammyS was saying is that you used the same symbol (T) for two different meanings (Torque and Tension)

"(Imagine holding something so that it just barely touches the ground. It will be touching the ground, but it's weight will be supported by your hand, so Fn would be zero. Same thing applies here, except Tsin(θ) only supports some of the weight, not all of it.)"

so then wouldn't my original answer still be correct then? Because we do not know the value of Fn and the only two components of the horizontal direction is the F and T cos theta in which we can isolated for T

Corey said:
so then wouldn't my original answer still be correct then? Because we do not know the value of Fn and the only two components of the horizontal direction is the F and T cos theta in which we can isolated for T

What about the static friction between the ground and the pole, isn't that another horizontal force?Edit:
Corey said:
Fy=0
Fn-mg-Tsin theta= 0

Another mistake is that Tsin(θ) should be positive (it's in the same direction as $F_n$)

So:
$F_n = mg - Tsin(θ)$

the question gave no indication of friction or any coefficients of kinetic or static friction so I do not think there would be any additional ground forces

Corey said:
the question gave no indication of friction or any coefficients of kinetic or static friction so I do not think there would be any additional ground forces

You don't need to know the coefficient of static friction as long as you assume that it's not being overcome.

Is it possible to solve the problem without static friction? You could choose the tension so that there's no rotation, but then there would be a net horizontal force and the beam would move (horizontally)

Or you could choose the tension so there's no net horizontal force, but then there would be a net torque and the beam would rotate.

Is it possible to have the beam not rotate and not move without static friction?

## What is static equilibrium?

Static equilibrium is a state in which the forces acting on an object are balanced, resulting in no net force and no acceleration. In other words, the object is at rest or moving at a constant velocity.

## What are the conditions for static equilibrium?

The conditions for static equilibrium are that the sum of all forces acting on the object must be equal to zero and the sum of all torques acting on the object must also be equal to zero. This means that both the forces and torques must be balanced in order for an object to be in static equilibrium.

## How do you calculate the net torque on an object?

The net torque on an object can be calculated by multiplying the force applied to the object by the distance from the pivot point. This can be represented by the equation T = F x r, where T is the torque, F is the force, and r is the distance from the pivot point.

## What are some real-world examples of static equilibrium?

Examples of static equilibrium in the real world include a book sitting on a table, a person standing still, or a bridge supporting the weight of cars and pedestrians. In each of these cases, the forces and torques acting on the objects are balanced, resulting in static equilibrium.

## How does static equilibrium differ from dynamic equilibrium?

Static equilibrium differs from dynamic equilibrium in that dynamic equilibrium occurs when an object is moving at a constant velocity, while static equilibrium occurs when an object is at rest or not moving. Additionally, in dynamic equilibrium, the forces and torques acting on an object may not be balanced, but they result in a net force of zero due to the object's constant velocity.

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