Tension in cable D is equivalent to the tension in cable B?

[yashboi123]

Homework Statement

In a repair shop a truck engine that has mass 429 kg
is held in place by four light cables (Figure 1). Cable A is horizontal, cables B
and D are vertical, and cable C makes an angle of 37.1∘
with a vertical wall. Tension in cable A is 757 N.

Relevant Equations

T - mg = ma

I found the tension of cable B by doing mg + Csin(37.1). I found C by doing 757(Tension in cable A) = Ccos(37.1).
I was just wondering if the tension in cable D is equivalent to the tension in cable B. If possible please show the steps on how you determined if they are equivalent or not.
Thank you!

 

[erobz]

Homework Statement: In a repair shop a truck engine that has mass 429 kg
is held in place by four light cables (Figure 1). Cable A is horizontal, cables B
and D are vertical, and cable C makes an angle of 37.1∘
with a vertical wall. Tension in cable A is 757 N.
Relevant Equations: T - mg = ma

View attachment 332527

I found the tension of cable B by doing mg + Csin(37.1). I found C by doing 757(Tension in cable A) = Ccos(37.1).
I was just wondering if the tension in cable D is equivalent to the tension in cable B. If possible please show the steps on how you determined if they are equivalent or not.
Thank you!

That sounds like part "C". per forum rules you must show us, if you get stuck after a reasonable attempt we help.

 

[yashboi123]

That sounds like part "C". per forum rules you must show us, if you get stuck after a reasonable attempt we help.

No it's not a part C, I'm just curious lol promise. Here is the full page

 

[erobz]

Well, if you want to find the tension in the rope $D$ do a free body diagram of the engine.

 

[yashboi123]

It would be tension upward(or normal force since they are equivalent in this situation) and mg downward. I suppose then they wouldn't be equivalent since in this situation we only take into account mg, not the vertical tension in cable C since cable D is below that point.

 

[yashboi123]

It would be tension upward(or normal force since they are equivalent in this situation) and mg downward. I suppose then they wouldn't be equivalent since in this situation we only take into account mg, not the vertical tension in cable C since cable D is below that point.

Is my thought process correct here?

 

[erobz]

Is my thought process correct here?

Yeah, there are two forces acting on the engine block on opposite directions, and the engine is not accelerating. We know one of them is it weight the other is the tension in rope $D$, hence;

$$T_D - mg = 0 $$

 

[yashboi123]

Yeah, there are two forces acting on the engine block on opposite directions, and the engine is not accelerating. We know one of them is it weight the other is the tension in rope $D$, hence;

$$T_D - mg = 0 $$

Thanks mate, that was a dumb mistake from me it's pretty clear since D is below C the vertical component of Tc wouldn't be considered.

 

[erobz]

Thanks mate, that was a dumb mistake from me it's pretty clear since D is below C the vertical component of Tc wouldn't be considered.

You're Welcome.