Find the magnitude, direction, and location of the force

[Inertialforce]

Homework Statement

The uniform bar shown below weighs 40N and is subjected to the forces shown. Find the magnitude, direction, and location of the force needed to keep the bar in equilibrium. (the L's in the diagram stand for "length")

Homework Equations

ΣFx , ΣFy , and ΣT (torque)

The Attempt at a Solution

The question says a force is needed to keep the bar in equilibrium so I used both translational and rotational equilibrium formulas for this question (because they are the only equilibrium formulas we have learned so far). I first went through the translational equilibrium calculations (ΣFx, and ΣFy) to find x and y components.

Translational Equilibrium:
ΣFy=may
ΣFy=0
50-60-70-40+(80sin(30))=0
50-60-70-40+(40)= -80
-80 = 0

therefore the unknown vertical component is = (+ or positive) 80N

ΣFx=max
ΣFx=0
-(80cos(30))=0
-69.2820323=0
-69 = 0

therefore the unknown horizontal component is = (+) 69N

Next I used the parallelogram method of vector addition to find the resultant and the answer I got for the resultant was 106N. I then went to go find the angle by using tan and the angle turned out to be 49(degrees). north of east.

Now that I have found the magnitude and direction of the unknown force, the only thing left to find is its location and this is where I am having trouble. I tried using the rotational equilibrium equation and this is what I got:

ΣT=0
T(counterclockwise)=T(clockwise)

(80sin(30))(l1 or lever arm 1) + 50(l3 or lever arm 3)= Mg(weight of beam)*(l2 or lever arm 2) + 60(l3 or lever arm 3) + 70(l3 or lever arm 3)

= (80sin(30))(1.00) + 50(0.2) = (40(0.5) + 60(0.2) + 70(0.2)

= 50 = 46

I have found this now but I am unsure as to how to use this to find the location of the unknown force needed to keep the bar in equilibrium.

 

[alphysicist]

Hi Inertialforce,

Homework Statement

The uniform bar shown below weighs 40N and is subjected to the forces shown. Find the magnitude, direction, and location of the force needed to keep the bar in equilibrium. (the L's in the diagram stand for "length")

Homework Equations

ΣFx , ΣFy , and ΣT (torque)

The Attempt at a Solution

The question says a force is needed to keep the bar in equilibrium so I used both translational and rotational equilibrium formulas for this question (because they are the only equilibrium formulas we have learned so far). I first went through the translational equilibrium calculations (ΣFx, and ΣFy) to find x and y components.

Translational Equilibrium:
ΣFy=may
ΣFy=0
50-60-70-40+(80sin(30))=0
50-60-70-40+(40)= -80
-80 = 0

therefore the unknown vertical component is = (+ or positive) 80N

ΣFx=max
ΣFx=0
-(80cos(30))=0
-69.2820323=0
-69 = 0

therefore the unknown horizontal component is = (+) 69N

Next I used the parallelogram method of vector addition to find the resultant and the answer I got for the resultant was 106N. I then went to go find the angle by using tan and the angle turned out to be 49(degrees). north of east.

Now that I have found the magnitude and direction of the unknown force, the only thing left to find is its location and this is where I am having trouble. I tried using the rotational equilibrium equation and this is what I got:

ΣT=0
T(counterclockwise)=T(clockwise)

(80sin(30))(l1 or lever arm 1) + 50(l3 or lever arm 3)= Mg(weight of beam)*(l2 or lever arm 2) + 60(l3 or lever arm 3) + 70(l3 or lever arm 3)

= (80sin(30))(1.00) + 50(0.2) = (40(0.5) + 60(0.2) + 70(0.2)

I don't think the values you are using here are correct. When you write out the toqure equation, you have to use the same pivot for the entire equation. The lever arms for the first three terms seem to be showing that the pivot is at the left end of the rod. However, the last two terms do not seem to have this. (For example, the fourth terms is suggesting that the 60 N force is 0.2m from the pivot.)

Also, I think it is incorrect to set the counterclockwise torques equal to the clockwise torques here, because the method you are using leaves out the unknown torque.

So you can do for the torques exactly what you did for the forces. For the forces, you found the net force of all the forces except for the one you are looking for, and then you knew the unknown force must be able to cancel that net force. Do you see how to apply this to the torques? (Of course there are other ways to set it up.)