Complicated Centre of Mass Question

[siemieniuk]

Hey all... I'm having trouble deciding where the heck I should even start with this question. Soooo... here it is (I've attached a picture of the diagram because the question wouldn't make any sense without it).

A non-uniform bar of weight W is suspended at rest in a horizontal position by two light ropes as shown in the following diagram. The angle one rope makes with the vertical is theta=36.9 degrees, the other makes the angle phi=53.1 degrees with the vertical. If the length L of the bar is 6.1 m, compute the distance x from the left hand end to the centre of gravity.

(see attachment)

Where I was headed, was to assign a weight to this bar, and calculate the tensions in the string. But, even if this was correct, I'm not sure where to go from there. Maybe I'm taking too much from a simple drawing, but would the fact that where the strings are suspended from on either side are at different heights have any effect on the outcome? Please advise...

Thanks,

Out of Equilibrium in Calgary

 

[Doc Al]

Treat this like any other equilibrium problem: Apply the conditions for equilibrium. The net force on the bar must be zero and the net torque (about any axis) must be zero. You can get three equations (x-components; y-components; torque) which will allow you to solve for the tension in each rope and "x" (the location of the center of mass).

 

[M98Ranger]

Hi there,

I can understand your frustration with this question. Finding the centre of mass of a non-uniform object can definitely be tricky, especially when it's suspended by multiple ropes. However, don't worry because I'm here to help!

First, you're on the right track by assigning a weight to the bar and calculating the tensions in the strings. This is a good starting point. To find the centre of mass, we will use the principle of moments, which states that the sum of the moments on an object is equal to zero.

In this case, we can take moments about the left end of the bar. The weight of the bar will create a clockwise moment, while the tensions in the strings will create counterclockwise moments. We can set up an equation like this:

(W * x) - (T1 * L * sin(theta)) - (T2 * L * sin(phi)) = 0

Where W is the weight of the bar, x is the distance from the left end to the centre of mass, T1 is the tension in the first string, T2 is the tension in the second string, and theta and phi are the angles given in the question.

Now, we have three unknowns (x, T1, T2) and only one equation. But don't worry, we have another piece of information that we can use to solve for x. Since the bar is at rest, we know that the sum of the vertical forces must be equal to zero. This means that the vertical components of the tensions in the strings must balance the weight of the bar. We can use this information to set up another equation:

T1 * cos(theta) + T2 * cos(phi) = W

Now we have two equations and two unknowns (T1 and T2). We can solve for these tensions and then plug them back into the first equation to solve for x.

As for the different heights of the suspending points, this will not affect the outcome because the tensions in the strings will adjust to keep the bar in equilibrium.

I hope this helps! Just remember to take your time and carefully consider all the forces acting on the bar. Best of luck to you!