Static Equilibrium and a uniform rod

[BillytheKid]

One end of a uniform 4.00-m long rod of weight Fg is supported by a cable. The other end rests against the wall, where it is held by friction. The coefficient of static friction between the wall and the rod is = 0.500. Determine the minimum distance x from point A at which an additional weight Fg (the same as the weight of the rod) can be hung without causing the rod to slip at point A. Angle cable and beam make is 36 degrees

I'm just staring at the problem we have done a lot in class involving the tension in the cable and how far one could walk before the beam breaks but that involves the beam being at a hinge on the wall. I have seen one only involving friction. I could use some hints

 

[Andrew Mason]

One end of a uniform 4.00-m long rod of weight Fg is supported by a cable. The other end rests against the wall, where it is held by friction. The coefficient of static friction between the wall and the rod is = 0.500. Determine the minimum distance x from point A at which an additional weight Fg (the same as the weight of the rod) can be hung without causing the rod to slip at point A. Angle cable and beam make is 36 degrees

I'm just staring at the problem we have done a lot in class involving the tension in the cable and how far one could walk before the beam breaks but that involves the beam being at a hinge on the wall. I have seen one only involving friction. I could use some hints

What you describe appears to be a lever with a fulcrum at the cable end. The issue is how much torque can be applied to this lever arm. The maximum downward force at A (contact point between rod and wall) without slipping is a function of the horizontal component of the force at A times \mu_s.

Does that help?

AM

 

[wais]

To solve this problem, we can use the concept of static equilibrium, which states that the forces acting on an object must be balanced for it to remain stationary. In this case, the forces acting on the rod are its weight (Fg) and the tension in the cable (T). We can also consider the friction force (Ff) acting at point A, which must be equal and opposite to the horizontal component of the tension in the cable.

First, let's draw a free body diagram of the rod:

At point A, the forces acting on the rod are the tension in the cable (T), the weight of the rod (Fg), and the friction force (Ff). The angle between the cable and the rod is 36 degrees, so the horizontal component of the tension (Tcos36) is equal to the friction force (Ff). We can also see that the vertical component of the tension (Tsin36) is equal to the weight of the rod (Fg).

Since the rod is in static equilibrium, the sum of the forces in the horizontal direction must be equal to zero:

Tcos36 - Ff = 0

Substituting in the value for the friction force (Ff = μFg), we get:

Tcos36 - μFg = 0

Similarly, the sum of the forces in the vertical direction must also be equal to zero:

Tsin36 - Fg = 0

Substituting in the value for the weight of the rod (Fg), we get:

Tsin36 - Fg = 0

Now, we can solve for the tension in the cable (T) by using the Pythagorean theorem:

T = √(Tcos36)^2 + (Tsin36)^2

Plugging in the values, we get:

T = √((μFg)^2 + (Fg)^2)

Next, we can use the fact that the additional weight (Fg) will cause the rod to slip at point A. This means that the friction force (Ff) must be equal to or greater than the horizontal component of the tension (Tcos36):

Ff ≥ Tcos36

Substituting in the value for the friction force (Ff = μFg) and the tension (T), we get:

μFg ≥ √((μFg)^2 + (Fg)^2)cos36