Static Equilibrium: holding a sign

[schwiffty]

Homework Statement

You are holding a sign as shown below. The sign (including the horizontal bar it hangs from) has a mass of 2.20 kg and is 40.5 cm wide. The sign is hanging from a 1.03 m tall, 4.46 kg vertical post. The sign is symmetric with a uniform mass distribution as implied by the drawing

What external force (+ if up, - if down) and torque (+ for CCW, - for CW) do you have to apply with your hand to keep the sign in static equilibrium? [Use g = 9.80 m/s2 and ignore the width of the vertical post.]

Homework Equations

Net Forces = 0

Torques (counter clockwise) = Torques (clockwise)

The Attempt at a Solution

I think that the only forces acting on the sign are vertical so for the net forces equation I have:

Forces up = Forces down =====> F_Hand = Mg + mg
Where M is the mass of the vertical post and m is the mass of the sign.

For the conservation of torques I have: d * F_Hand = d*Mg + d*mg
Where d is the lever arm in meters.

(The axis of rotation is at the left side.)

0.405 * F_Hand = 0.405 * (4.46kg*9.8) + 0.2025 * (2.20kg*9.8)F_Hand = (17.7017 N*m + 4.3659 N*m) / 0.405 m

F_Hand = 54.488 N Torque = 22.0676 N*m

This answer is wrong and I do not know why.

 

[berkeman]

Homework Statement

You are holding a sign as shown below. The sign (including the horizontal bar it hangs from) has a mass of 2.20 kg and is 40.5 cm wide. The sign is hanging from a 1.03 m tall, 4.46 kg vertical post. The sign is symmetric with a uniform mass distribution as implied by the drawing

What external force (+ if up, - if down) and torque (+ for CCW, - for CW) do you have to apply with your hand to keep the sign in static equilibrium? [Use g = 9.80 m/s2 and ignore the width of the vertical post.]

Homework Equations

Net Forces = 0

Torques (counter clockwise) = Torques (clockwise)

The Attempt at a Solution

I think that the only forces acting on the sign are vertical so for the net forces equation I have:

Forces up = Forces down =====> F_Hand = Mg + mg
Where M is the mass of the vertical post and m is the mass of the sign.

For the conservation of torques I have: d * F_Hand = d*Mg + d*mg
Where d is the lever arm in meters.

(The axis of rotation is at the left side.)

0.405 * F_Hand = 0.405 * (4.46kg*9.8) + 0.2025 * (2.20kg*9.8)F_Hand = (17.7017 N*m + 4.3659 N*m) / 0.405 m

F_Hand = 54.488 N Torque = 22.0676 N*m

This answer is wrong and I do not know why.

Your vertical force calculation looks okay, but I'm not understanding your torque equations. The vertical post itself would not require a torque to keep vertical. What are you taking as the mass and center of mass for the torque from the sign?

 

[TSny]

Hello, Welcome to PF!

What external force (+ if up, - if down) and torque (+ for CCW, - for CW) do you have to apply with your hand to keep the sign in static equilibrium? [Use g = 9.80 m/s2 and ignore the width of the vertical post.]

Note that the hand applies both a force and a torque. The torque is in the form of a "couple". The torque due to a couple is independent of the choice of location of the axis of rotation.

I think that the only forces acting on the sign are vertical so for the net forces equation I have:

Forces up = Forces down =====> F_Hand = Mg + mg
Where M is the mass of the vertical post and m is the mass of the sign.

Does this give the correct answer for F_Hand?

For the conservation of torques I have: d * F_Hand = d*Mg + d*mg
Where d is the lever arm in meters.

(The axis of rotation is at the left side.)

You did not include the torque (couple) provided by the hand.
I recommend taking the axis of rotation to be some point on the vertical rod.

 

[schwiffty]

Hello, Welcome to PF!

Note that the hand applies both a force and a torque. The torque is in the form of a "couple". The torque due to a couple is independent of the choice of location of the axis of rotation.Does this give the correct answer for F_Hand? You did not include the torque (couple) provided by the hand.
I recommend taking the axis of rotation to be some point on the vertical rod.

Hi!

I can't find out if my answer for the force of the hand is correct because the question requires both parts to be entered to tell me if its right.

When you say the torque is in the form of a couple, does this mean that there is some horizontal force acting on the vertical bar I need to solve for?

 

[TSny]

Hi!

I can't find out if my answer for the force of the hand is correct because the question requires both parts to be entered to tell me if its right.

I see.

When you say the torque is in the form of a couple, does this mean that there is some horizontal force acting on the vertical bar I need to solve for?

See figure below. The couple can be thought of as due to the two red forces f. You do not need to find f. You just need to find $\tau_H$.

 

[schwiffty]

Turns out I just needed to use the vertical post/hand as the axis of rotation, giving me a torques equation of:

0.2025 * 2.20kg * 9.8 = Torques CW

(torque caused by the sign) = (torque provided by the hand to keep the system in static equilibrium)

... giving me an answer of -4.3659, negative since its clockwise.

Thanks TSny and berkeman.