As shown the truss is loaded by the forces p1 191 lb and p2...

As shown, the truss is loaded by the forces $P_1 = 191$ lb and $P_2 = 400$ lb and has the dimension $a = 2.40$ ft. Using the method of joints, determine $F_{AB}$, $F_{BC}$, and $F_{BD}$, the magnitude of the force in each of the members connected to joint B. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption. Express your answers numerically in pounds to three significant figures separated by commas. View Available Hint(s) $F_{AB}$, $F_{BC}$, $F_{BD}$ = lb Submit

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00:01 So therefore, the sum of our forces in the x direction are going to equal to zero, and then the sum of our forces in a y direction are also equal to zero.

00:07 So if we look at the sum of our forces in the x direction, then we're going to have to be a of x plus d of x equal to 0.

00:18 So, and then if we look at the forces in the y direction, then we're going to have d of y minus p1 minus p2, to equal zero.

00:32 So therefore the sum of our forces in the x direction are going to equal to zero and then the sum of our 40, then we get that dy is going to equal to 980 pounds.

00:43 And now if we look at the sum of our moments, some in the moments around a, then we're gonna have three times dx minus four times this would be p2 equal to zero.

01:01 Solving for this we'll have dx equal to negative 420 times 4 over 3 so therefore the sum of our forces in x -direction are going to equal to 0 and then it's 20 so that's going to give us 560 so therefore to sum about forces in x -direction are going to equal to 0 negative 560 so then so therefore the sum about forces in x -direction are going to equal to 0 and solve for a x and we'll have that x is also going to be or is it going to be positive 560 since d x is negative 560 so now if we look at joint b we look at the sum of our forces in the x direction there we'll get that the force from b to a is equal to zero and then the sum of our forces for y at joint b is going to be f f bc and that's going to be equal to just p2 which is just 420.

02:16 So now if we look at joint d, let's look at the forces there.

02:22 Well, this is going to be just f of dc minus 560 equal to zero...

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