Calculating Moment of Force in a Right Triangle - 3-4-5 Triangle Example

[SpringMorning]

Homework Statement

I have a 3-4-5 right triangle with the 3 as the base and the 5 as the hypotenuse.

The corner of the 3 and the 5 is point A

The corner of the 3 and the 4 is point B

At B, perpendicular to the base is a force of 100lb going down- into the corner. (Fv)

At B, Parallel to the base is a force of 100lb going "into" the corner. (Fh)

Homework Equations

The question is asking: Find the moment of the force with respect to A.

The Attempt at a Solution

I am assuming the angles are 53 deg at A, 37 deg at B (and the 90 of course).

I am treating this like a torque equation but since the forces are going into the figure. I have no idea where to start. Any help would be greatly appreciated. It's been about 5 yrs since I have done any physics. I am taking a fluid dynamics class to prepare to begin getting my MS in Environmental Engineering (Water) in the fall.

Thank you!

 

[SpringMorning]

Homework Statement

I have a 3-4-5 right triangle with the 3 as the base and the 5 as the hypotenuse.

The corner of the 3 and the 5 is point A

The corner of the 3 and the 4 is point B

At B, perpendicular to the base is a force of 100lb going down- into the corner. (Fv)

At B, Parallel to the base is a force of 100lb going "into" the corner. (Fh)

Homework Equations

The question is asking: Find the moment of the force with respect to A.

The Attempt at a Solution

I am assuming the angles are 53 deg at A, 37 deg at B (and the 90 of course).

I am treating this like a torque equation but since the forces are going into the figure. I have no idea where to start. Any help would be greatly appreciated. It's been about 5 yrs since I have done any physics. I am taking a fluid dynamics class to prepare to begin getting my MS in Environmental Engineering (Water) in the fall.

Thank you!

I am so incredibly embarrassed... I had my calculator set on radians, instead of degrees. But please check my answer if you would... I got 100 lb ft (Yes the english units are driving me nuts too but it's the way the teacher did it... when oh when will the US catch up to the rest of the world when it comes to the metric system!)

 

[nvn]

SpringMorning: I am currently unsure what the dimensions of the triangle sides are. I understand the triangle proportions; but no actual dimensions with units are stated. Nonetheless, the moment about point A would be the applied force multiplied by the perpendicular distance from point A to the force. I currently did not understand how you obtained your answer. If you show your work, someone might check your math.

 

[SpringMorning]

SpringMorning: I am currently unsure what the dimensions of the triangle sides are. I understand the triangle proportions; but no actual dimensions with units are stated. Nonetheless, the moment about point A would be the applied force multiplied by the perpendicular distance from point A to the force. I currently did not understand how you obtained your answer. If you show your work, someone might check your math.

I am sorry for not being more clear.

The dimensions are in feet and lbs.

My work: T=rFsin(theta)

T1=5 ft*100lbs*sin(37)
T2=5 ft*100lbs*sin(53)

Sum T= T2-T1=399.318-399.318=approx 100 (98.41...)

The triangle is 3 ft base, 4 foot high, 5 ft hyp

Forces Fvertical= 100 lbs (perp to base)
Fhorizontal=100 lbs (parallel to base)

Both forces come into the corner of the hyp and height (5 and 4)

I hope this helps.

Thank you!

 

[nvn]

My current understanding is, the (x, y) coordinates, in units of feet, of your triangle vertices are A(0, 0), B(3, 4), C(3, 0). And the forces applied to point B are Fh = -100 lbf, and Fv = -100 lbf. If this is correct, then your answer is correct. Please correct me if I am misunderstanding your diagram. See also this https://www.physicsforums.com/showpost.php?p=2191084".

 

[SpringMorning]

My current understanding is, the (x, y) coordinates, in units of feet, of your triangle vertices are A(0, 0), B(3, 4), C(3, 0). And the forces applied to point B are Fh = -100 lbf, and Fv = -100 lbf. If this is correct, then your answer is correct. Please correct me if I am misunderstanding your diagram. See also this https://www.physicsforums.com/showpost.php?p=2191084".

Yes, your understanding is correct.

Thank you for checking this for me.

I have a feeling that this forum is going to be a lifesaver for me the rest of this semester!

Thank you so much!