Is there anyone with the right answer to this!!!!! As shown, a roller at point A

Question

Is there anyone with the right answer to this!!!!!
As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 18.0kg . The beam is subjected to the forces F1 = 58.0N and F2 = 73.0N . The dimensions are l1 = 0.300m and l2 = 2.60m . (Figure 2) What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible. FIND Fa & Fb Express your answers numerically in newtons to three significant figures separated by a comma. As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 18.0kg . The beam is subjected to the forces F1 = 58.0N and F2 = 73.0N . The dimensions are l1 = 0.300m and l2 = 2.60m . (Figure 2) What are the magnitudes FA and FB of the reaction forces FA and FB at points A and B, respectively? The beam's height and width are negligible. FIND Fa & Fb Express your answers numerically in newtons to three significant figures separated by a comma.

Explanation / Answer

force due to weight of rod=18.0 kg * 9.8 = 176.4N of force at (.3+2.6)/2 = 1.45m

trying to solve for the normal force at A, NA**anticlockwise = positive**
(58N * 2.6m) + (176.4N * 1.45m) - (2.9m * NA* 4/5) = 0
NA= 175.25N

solving for BX
sum of forces in x direction = 0; **positive direction to the right**
-BX- 73sin(15) + (NA*3/5) = 0
BX= 86.256N

solving for BY
sum of forces in the y direction = 0l **positive direction is upwards**
-176.4 - 58 - 73.cos(15) + (NA* 4/5) + BY= 0
BY= 164.71N

so as the question asks, FA= 175.25N and;
FB=

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