What is the tension in the cable and the reaction at D?

[yaro99]

Homework Statement

A force P of magnitude 90 lb is applied to member ACE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D.

Homework Equations

ƩM=0
ƩF_x=0
ƩF_y=0

The Attempt at a Solution

I am stuck on part a). Here is my equilibrium equation:
ƩM_D=90*9 + T*3 - (5/13)*(9)*T - (12/13)*(7)*T = 0

Solving for T I get the wrong answer.

 

[voko]

Perhaps you could explain each term in the moments equation.

 

[yaro99]

Perhaps you could explain each term in the moments equation.

90*9: 90 is the P force, and 9 is the perpendicular distance from C to D (12-3)

T*3: This is for the portion of the cable running from B to E. Perpendicular distance is a=3.

(5/13)*(9)*T: horizontal component of the AB portion of the cable. I got 5/13 from the ratio of the triangle (hypotenuse is 13 by pythagorean theorem). 9 is the perpendicular distance.

(12/13)*(7)*T: vertical component of the AB portion of the cable. Same triangle ratio, with 7 as the perpendicular distance.

 

[voko]

Let's look at the force applied at A. What is the direction and the magnitude of the lever arm relative to D?

 

[yaro99]

Let's look at the force applied at A. What is the direction and the magnitude of the lever arm relative to D?

I thought it was the AB tension from the last two terms in my equation. I don't see any other forces acting on A.

edit: Its direction would be 67.4° from the horizontal, and the magnitude would be whatever T is.

 

[voko]

You are talking about the magnitude and the direction of force. I am talking about the lever arm: this is the distance from the point about which moments are calculated to the point to which the force is applied.

I notice that you did in fact decomposed the lever arm into horizontal and vertical components, which is almost correct. Almost, because I think you neglected the signs. Which is why it is important to know what the direction of the lever arm is.

 

[yaro99]

You are talking about the magnitude and the direction of force. I am talking about the lever arm: this is the distance from the point about which moments are calculated to the point to which the force is applied.

I notice that you did in fact decomposed the lever arm into horizontal and vertical components, which is almost correct. Almost, because I think you neglected the signs. Which is why it is important to know what the direction of the lever arm is.

Wouldn't it be different for both components? For the vertical T component, it goes to the right from A to C (+7in). For the horizontal component, it goes up from C to D (+12in).

 

[voko]

Actually, the directions are exactly the opposite of what you said. Recall that the vector of the displacement must be taken from D, not toward D. Despite this, I now think your original equation is correct. What is value of tension that you get from it? Why do you think it is not correct?

 

[yaro99]

Actually, the directions are exactly the opposite of what you said. Recall that the vector of the displacement must be taken from D, not toward D. Despite this, I now think your original equation is correct. What is value of tension that you get from it? Why do you think it is not correct?

Whoops, it looks like I did get it right the first time, I just did the calculations wrong. I did it again and got the right answer. Sorry about that!