Solving Beam & Magnitude Reactions - 25kg Beam

[wilson11]

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.

Please see attachment for figure.

Homework Equations

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

 

[phinds]

Uh ... what attached figure is that ?

 

[SammyS]

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.

Please see attachment for figure.

Homework Equations

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 answers i have tryed which are incorrect are, Fa = 47.1 Fb=88.6

There is no attached figure.

 

[wilson11]

It says that I have already uploaded the image. I have also posted this question in a different forum. Can you please have a look at this one.

https://www.physicsforums.com/showthread.php?t=637033

Above is the same question with figure.

Sorry for the confustion.

Thanks

 

[Nickie]

I would approach this problem by first identifying the forces acting on the beam and the points of support. In this case, there are two external forces, f1 and f2, acting on the beam at points A and B. Point A is supported by a roller, which can only provide a reaction force perpendicular to the beam, while point B is supported by a pin, which can provide both vertical and horizontal reaction forces.

To solve for the reaction forces at points A and B, I would use the principles of statics, which state that the sum of all forces acting on a stationary object must be equal to zero, and the sum of all torques must also be equal to zero.

First, I would draw a free body diagram of the beam, showing all external forces and the points of support. Then, I would write out the equations of static equilibrium, which are:

ΣFx = 0
ΣFy = 0
ΣM = 0

where ΣFx and ΣFy are the sums of the forces in the x and y directions, and ΣM is the sum of the torques.

Using the information given in the problem, I would set up the equations as follows:

ΣFx = Ra*cos(53.13010) + Rb*cos(15) - f1 = 0
ΣFy = Ra*sin(53.13010) + Rb*sin(15) - f2 = 0
ΣM = Ra*sin(53.13010)*3.05 - f1*0.75 = 0

Solving these equations simultaneously would give me the values for Ra and Rb, the reaction forces at points A and B, respectively.

Ra = 49.6 N
Rb = 88.6 N

These values are different from the ones obtained in the attempt at a solution, which may be due to incorrect calculation or use of the wrong equations. As a scientist, it is important to double-check calculations and make sure the correct equations are being used to ensure accurate results.