What is the Maximum Distance for the Load on the Rod?

[hs764]

Homework Statement

A uniform rod AB of length 5 m and weight 50 N is pivoted at A and held in equilibrium by a rope as shown. A load of 100 N hangs from the rod at a distance x from A. If the breaking strength of the rope is 50 N, find the maximum value of X.

Homework Equations

τ_net = 0, τ = Fr⊥[/B]

The Attempt at a Solution

The solution I came up with is τ = [50(sin(37))*5(cos(37)) - (50*(5(cos(37)))/2) - (100(cos(37)x)) = 0, which gave me x = 0.29 m. I just wanted to verify that I'm doing this correctly. [/B]

 

[SammyS]

Homework Statement

A uniform rod AB of length 5 m and weight 50 N is pivoted at A and held in equilibrium by a rope as shown. A load of 100 N hangs from the rod at a distance x from A. If the breaking strength of the rope is 50 N, find the maximum value of X.

View attachment 81047

Homework Equations

τ_net = 0, τ = Fr⊥[/B]

The Attempt at a Solution

The solution I came up with is τ = [50(sin(37))*5(cos(37)) - (50*(5(cos(37)))/2) - (100(cos(37)x)) = 0, which gave me x = 0.29 m. I just wanted to verify that I'm doing this correctly. [/B]

Could you show us the major steps that led you to that solution?

 

[hs764]

I just calculated the torque for the rope, the rod, and the block. For the ropeI got τ = 50*sin(37)*5*cos(37), for the rod I got τ = 50*(5*cos(37)/2), and for the block I got τ = 100*cos(37)*x. Then I just plugged that into τ_rope - τ_rod - τ_block = 0. I think I was slightly off the first time because I calculated it again and got x = 0.25 m.

 

[SammyS]

I just calculated the torque for the rope, the rod, and the block. For the ropeI got τ = 50*sin(37)*5*cos(37), for the rod I got τ = 50*(5*cos(37)/2), and for the block I got τ = 100*cos(37)*x. Then I just plugged that into τ_rope - τ_rod - τ_block = 0. I think I was slightly off the first time because I calculated it again and got x = 0.25 m.

I this the torque about the pivot, A ?

The following is wrong.

"For the ropeI got τ = 50*sin(37)*5*cos(37)"​

That's only the torque due to the vertical component of the tension in the rope. The horizontal component of the tension in the rope also produces torque about point A.

It might be easier to find the component of the tension in the rope that's perpendicular to the rod.

 

[hs764]

Right, because it's the horizontal component of the tension that produces torque here. So this way I got τ_tension = (50)*sin(53)*3.01 m, which gave me x = 0.26 m.

 

[SammyS]

Right, because it's the horizontal component of the tension that produces torque here. So this way I got τ_tension = (50)*sin(53)*3.01 m, which gave me x = 0.26 m.

The horizontal component of what force produces the torque here (where)?

You need to be much more specific in answering.

If you are splitting the tension in the rope, T, into horizontal & vertical components, you need to recognize that each of these components produce torque about point A, so you need to include both contributions in calculating the overall torque.

(As I suggested previously, you can find the overall torque that the tension in the rope, T, produces about point A, by using the component of T which is perpendicular to the rod.)​

 

[hs764]

Ahhhh okay I was thinking about the axis incorrectly. So the total torque produced by the tension in the rope would be (50 N)*(cos 37)*(3.01 m) + (50 N)*(sin 37)*(3.99 m). Alternatively, I could just use (50 N)*(5 m)*(sin 74), right? And then solving for x gives x = 1.76 m?

 

[SammyS]

Ahhhh okay I was thinking about the axis incorrectly. So the total torque produced by the tension in the rope would be (50 N)*(cos 37)*(3.01 m) + (50 N)*(sin 37)*(3.99 m). Alternatively, I could just use (50 N)*(5 m)*(sin 74), right? And then solving for x gives x = 1.76 m?

Yes.

That looks good !

 

[hs764]

Awesome, thank you!

 

[Mkbul]

Hello guys. I attempted a solution to this problem as well, because it looked interesting since i am giving exams. I got a different result than you folks though. Here is my solution:

EDIT #3: I shouldn't be doing this late at night. Here is my final attempt. Not sure if right or wrong, but now i really can't spot any mistake.

 

[SammyS]

Hello guys. I attempted a solution to this problem as well, because it looked interesting since i am giving exams. I got a different result than you folks though. Here is my solution:

EDIT #3: I shouldn't be doing this late at night. Here is my final attempt. Not sure if right or wrong, but now i really can't spot any mistake.

Do the calculation. I think it's the same result O.P. finally got.

 

[hs764]

Yeah, I did your calculation and still got x = 1.76 m.