Roller at point A, Magnitudes

Homework Statement

As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 25.0 kg . The beam is subjected to the forces f1 = 50.0N and f2= 79.0N . The dimensions are l1= 0.750m and l2= 2.30m . (Figure 2) What are the magnitudes and of the reaction forces and at points A and B, respectively? The beam's height and width are negligible.

The Attempt at a Solution

(79*cos(15)*3.05)+(50*0.75) = Rb *3.05
Rb=88.6

Ra * sin(53.13010)*3.05 = 50*2.3
Ra=47.1311

I am not sure if this is how you do this type of question..?
Also the answers are not correct as when I go to see if they are correct they come back as incorrect. The ansers i have tryed which are incorrect are, Fa = 47.1 Fb=88.6

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PhanthomJay
Homework Helper
Gold Member
you forgot to include the weight from the beam , the resultant of which acts at its center. Also, what you are calling and solving for Rb is actually Rb_y...the vert comp of the support (reaction force)force at the pin B.

Sorry so what would the equation look like now (with the weight)?
Where do i put it?

Thanks

PhanthomJay
Homework Helper
Gold Member
When determining force reactions, the weight of the uniform beam may be represented by a single weight force acting at its center of gravity (its mid point).

(79*cos(15)*3.05)+(50*0.75) = Rb *3.05
Rb=88.6

Ra * sin(53.13010)*3.05 = 50*2.3
Ra=47.1311
I may be very slow, but i do understand what you are saying, however i am haveing trouble to where in this equation(above) does the force from the beam go. Do i just times the weight by 9.81 to find N and then just add or what?

Thanks

PhanthomJay
Homework Helper
Gold Member
The mass of the beam in kg times 9.8 gives you the weight of the beam in Newtons. Place this downward force at 3.05/2 m from one end. Then redo your moment equations, which otherwise appear correct, except Rb should be Rb_y.

(79*cos(15)*3.05)+(50*0.75)*25*9.81= Rb_y*3.05

Is that correct above?

PhanthomJay
Homework Helper
Gold Member
(79*cos(15)*3.05)+(50*0.75)*25*9.81= Rb_y*3.05

Is that correct above?
No-o. The moment from the 79 N force about A is correct. The moment of the 50 N force about A is 50 * 0.75. The moment of the weight force about A is 25*9.8 * (___???___). Add all three moments up and set them equal to Rb_y(3.05). Solve for Rb_y and continue...

Sorry if I seem very slow, But is what you are saying is:
(79*cos(15)*3.05)+(50*0.75)+(25*9.81)= Rb_y*3.05

Please correct me if I am still getting it wrong..

PhanthomJay
Homework Helper
Gold Member
Sorry if I seem very slow, But is what you are saying is:
(79*cos(15)*3.05)+(50*0.75)+(25*9.81)= Rb_y*3.05

Please correct me if I am still getting it wrong..
You correctly multiplied the other forces by the moment arm distances to A to find the value of the moments about A...but you are forgetting to multiply the weight force by its moment arm....the moment arm of the weight force about A is the perpendicular distance from its line of action to A....which is how much ??

(79*cos(15)*3.05)+(50*0.75)+(25*9.81*0.75)= Rb_y*3.05

I think i read that right. Is that what you mean? ^^^^

PhanthomJay
Homework Helper
Gold Member
(79*cos(15)*3.05)+(50*0.75)+(25*9.81*0.75)= Rb_y*3.05

I think i read that right. Is that what you mean? ^^^^
You are still not getting it right...that's OK, it takes awhile.... The weight resultant force of 25(9.8) N acts at the center of the beam, at 3.05/2 = 1.525 m from A....so why are you using 0.75 m as the moment arm of the weight force about A when you should be using __???___ m

(79*cos(15)*3.05)+(50*0.75)+(25*9.81*1.525)= Rb_y*3.05

I am sorry If i am wasteing your time, But is the above what you mean? I am sorry to be so slow but I am very confused about this question.

PhanthomJay