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

[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?

 

[Nathanael]

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)

 

[Nathanael]

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

 

[Corey]

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

 

[Corey]

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

 

[Nathanael]

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.)

 

[SammyS]

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 ?

 

[Corey]

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

 

[Nathanael]

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)

 

[Corey]

"(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

 

[Nathanael]

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:

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)

 

[Corey]

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

 

[Nathanael]

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?