Easy statics problem -- Block being hoisted by a pulley and rope

[Oxford365]

I drew a FBD about point A and came up with the two equations:
∑Fx=0 20sinθ - T-ab(cos70)=0
ΣFy =0 -20cosθ + T-ab(sin70) -20=0

My answer is not coming out so I either messed up on the FBD or the equations, some help would be appreciated. I'm on 3-10 by the way.

 

[BvU]

Hard to distinguish what your equations are about; what is the 20 ?
And with T - ab you mean T_ab ?
How come W does not feature ? nor the 80 lb ? (lbf ?)

 

[SteamKing]

Easy statics problem ... which is why it belongs in the Intro Physics HW forum and not one of the technical engineering forums.

 

[Oxford365]

Hard to distinguish what your equations are about; what is the 20 ?
And with T - ab you mean T_ab ?
How come W does not feature ? nor the 80 lb ? (lbf ?)

Yes that is what I mean. I figured the force at D is 20lb because the block has 20lb of force and the cable holding it has 20lb of tension so D would as well (same cable). So I tried to resolve that 20lbs into the x and y direction.

 

[BvU]

Sorry, only saw 3-11.
Equations look good. Must be the solution part. What do you do to solve $$
\quad 20\sin\theta = T_{AB}\sin(20^\circ) \\
\ 20+ 20 \cos\theta = T_{AB}\cos(20^\circ) \quad {\rm ? }$$

 

[NickAtNight]

I figured the force at D is 20lb because the block has 20lb of force and the cable holding it has 20lb of tension so D would as well (same cable).

So then your first equation was actually $$ F_c = F_f $$ where $$ F_c = 20 lbs. $$ So $$ F_f = 20 lbs. $$
You probably should have written that equation.

 

[Oxford365]

Sorry, only saw 3-11.
Equations look good. Must be the solution part. What do you do to solve $$
\quad 20\sin\theta = T_{AB}\sin(20^\circ) \\
\ 20+ 20 \cos\theta = T_{AB}\cos(20^\circ) \quad {\rm ? }$$

I just try solving like any other system of equations. 2 unknowns and 2 equations which is leaving me with funky answers.

 

[NickAtNight]

What do you get if you solve for the extremes? Sin and cos vary between 0 and 1. You should be able to bracket your answer.

 

[NickAtNight]

You might as well revise the equations to solve for Tab. Consider replacing the fixed signs with their numbers and divide thru.

 

[NickAtNight]

Hint: you have a formula error... Sin is hypotenuse over the what? Cos is hypotenuse over the what?

hint2: one of the angles is taken from the horizontal x axis. The other is from the vertical y axis. How does this affect if you use sin or cos...

 

[Oxford365]

I figured it out guys, thanks for the help.

 

[NickAtNight]

I figured it out guys, thanks for the help.

And the answer is what? I got 31.32 with the fixed equations.

 

[Oxford365]

And the answer is what? I got 31.32 with the fixed equations.

The angle is 40degrees and Tab=37.6lb

 

[NickAtNight]

Really, I am getting degrees =34.91 and Tab 33.46? Original equation above did not divide by cos(20)

Hmmm, not very good at using Numbers on an iPad yet. Perhaps I should try it by hand.

Okay. got it. Tab = 37.6 and angle = 40.

 

[NickAtNight]

I figured it out guys, thanks for the help.

Edit: Add in tension equation

Alternative solution. Rotate your x/y axis so that x is parallel to $$ T_{ab} $$

A-C is 20 degree angle from axis.
A-D is unknown angle

$$ F_ {ac} = F_{ad} $$

$$ F_{ac} cos \theta = F_{ad} cos (uk) $$
And therefore uk angle = theta.

And

$$ F_{ab} = F_{ac} sin \theta + F_{ad} sin (uk) $$

which reduces to

$$ F_{ab} = 2 * F_{ac} sin \theta $$
$$ F_{ab} = 2 * 20 sin (20 degrees) $$

Voila