# Rotational Equilibrium uniform boom problem

• JinM
In summary, the boom is supported by a cable, and a weight hangs from its top. The cable applies a force to the bottom of the boom, and the weight creates a reaction force on the bottom of the boom.
JinM

## Homework Statement

(Figure is attached)
A 1200.0 N uniform boom is supported by a cable,as
shown in Figure 8-22.The boom is pivoted at the bot-
tom,and a 2000.0 N weight hangs from its top.Find the
force applied by the supporting cable and the compo-
nents ofthe reaction force on the bottom ofthe boom.
(See Sample Problem 8B.)

Legend:
t denotes torque
T denotes tension
d denotes distance from pivot
Fg denotes gravitational force
Rx denotes horizontal component of reaction force
Ry denotes vertical component of reaction force.

## Homework Equations

tnet = Fd sin theta
Fnet =ma

## The Attempt at a Solution

Take point A as pivot
tnet = 0
tnet = tT + tFg1 + tFg2 = 0

+(T x 0.75L sin 90) - (Fg1 x 0.5L sin 25) - (Fg2 x L sin 25) = 0

T = 1464 N

Fnet = T + Fg1 + Fg2 + Rx + Ry = 0

Fx = Fg1 cos 205 + Fg2 cos 205 + Rx = 0
Rx = 2912 N

Fg = Ry + Fg1 sin 205 + Fg2 sin 205 + T sin 90 = 0
Ry = - 111

R = Sqrt(-111^2 + 2912^2)
R = 2915.7 N

My angles for T, Fg1, and Fg2 are 90, 205, and 205, respectively. I got this wrong, and I'm curious why.

#### Attachments

• Picture 2.png
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The weights act vertically.

How are you using x and y?

I took the forces perpendicular to the inclined boom, for torques ask for the forces to be perpendicular to the lever arm. I don't see why weights should act vertically, though. Shouldn't it be perpendicular to the boom, as I've done above?

Which way does gravity act?

downwards, but doesn't it have a component that is perpendicular to the boom?

Sure.

So in a sense, I am correct in taking the angles as they are above? I only ask this because my teacher took tension T at 155 degrees and F_g at 270 degrees, which doesn't make sense to me!

Your teacher is measuring angles with respect to the x-axis (horizontal), which is standard practice.

We got different solutions though -- I wish it was just a matter of convention.

JinM said:
We got different solutions though -- I wish it was just a matter of convention.
When I get a few minutes, I'll look through your work. But since you use different x and y axes, could that be the problem? Aren't you supposed to give horizontal and vertical components?

I think that is what I'd already done? Maybe it's a little messy up there -- it could definitely be redone more efficiently. For one thing, since tension is perpendicular, it's just simply positioned at 90 degrees to the boom. F_g would be in the opposite direction at 25 degrees, so the component would be F_g*sin(25). See, my reference is the boom -- that's where my horizontal is, and that is why, I think, our solutions differ.

I'd appreciate it if you could look through my work!

OK, I looked it over. Your work and angles look OK, given your choice of axis. (Except for some slight arithmetic differences.) Of course, if you don't translate your reaction force components into the form asked for (horizontal and vertical), then of course you will have different answers.

(If someone's grading your work, you must make it easy for them to figure out what you're doing. You should state up front that your x and y-axis are not horizontal and vertical, but parallel and perpendicular. But still, it's up to you to answer the question as asked.)

## 1. What is rotational equilibrium in a uniform boom problem?

Rotational equilibrium in a uniform boom problem occurs when the forces acting on an object are balanced, causing the object to remain in a state of rest or constant rotation.

## 2. How is rotational equilibrium achieved in a uniform boom problem?

In order to achieve rotational equilibrium in a uniform boom problem, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. This ensures that there is no net torque acting on the object.

## 3. What factors affect rotational equilibrium in a uniform boom problem?

The length and weight of the boom, as well as the location of the pivot point, can all affect rotational equilibrium in a uniform boom problem. Additionally, the magnitude and direction of the forces acting on the boom can also impact rotational equilibrium.

## 4. How can we calculate the forces needed for rotational equilibrium in a uniform boom problem?

To calculate the forces needed for rotational equilibrium in a uniform boom problem, we can use the equation Στ = 0, where Στ represents the sum of the torques and must equal zero for equilibrium to be achieved. We can also use the principles of vector addition and trigonometry to determine the necessary forces.

## 5. What are some real-life applications of the uniform boom problem and rotational equilibrium?

The concept of rotational equilibrium in a uniform boom problem can be applied to various real-life situations, such as balancing a see-saw or a ladder against a wall. It is also important in engineering and construction, as it helps ensure the stability and safety of structures such as cranes and bridges.

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