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Problem 4 An object of mass m=100-kg is suspended from joint D of the truss. The structure is supported with a pin support at A and a roller at B. Draw the free-body diagram of the truss and calculate the reaction forces (in kN). Then, using the Method of Joints, determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C).

          Problem 4
An object of mass m=100-kg is suspended from joint D of the truss. The structure is supported with a pin support at A and a roller at B. Draw the free-body diagram of the truss and calculate the reaction forces (in kN). Then, using the Method of Joints, determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C).
        
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Problem 4
An object of mass m=100-kg is suspended from joint D of the truss. The structure is supported with a pin support at A and a roller at B. Draw the free-body diagram of the truss and calculate the reaction forces (in kN). Then, using the Method of Joints, determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C).

Added by Dana R.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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An object of mass m = 100 kg is suspended from joint D of the truss. The structure is supported with a pin support at A and a roller at B. Draw the free-body diagram of the truss and calculate the reaction forces (in kN). Then, using the Method of Joints, determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C).
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Transcript

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00:01 Here we have a static equilibrium problem, and so we need to start thinking with a free body diagram.
00:07 So we can take our body free from the external reactions and replace those reactions with the forces.
00:13 So we have a, b, c, that is c, and this is d.
00:21 We have our applied loaded d, which is equal to 100 kilograms times g.
00:26 The acceleration to grab down that gives us the weight.
00:28 Now at point b we have a roller, which means we can only exist a vertical load, and we'll use a normal reaction, so this is going to be b of y.
00:40 And at a, we have a pin, which can resist no moment, but forces in both directions.
00:46 So we have an ay and an ax.
00:51 So we need to pick an axis, and we've already shown which direction it is.
00:55 We'll pick x and y.
00:56 We'll put the origin at a, and that's going to be helpful in part a, to find the external reactions.
01:07 So what we want to do then is look at, well we have part a, just the first part, find the external reactions.
01:16 And so if the body is in static equilibrium, that means forces as well as moments, summed up equals zero on the body.
01:24 We take positive to be counterclockwise.
01:26 Now let's take them about point a.
01:28 We know that a moment is a force times a distance, if those are perpendicular, distance goes to zero, so does moment.
01:36 So the forces at a make no moment about a because they go directly through that point.
01:40 So the only unknown we have in the moment equation about a is b .y.
01:45 So that's good.
01:46 We have one equation and one unknown we can solve for.
01:48 If we see b .y, that tends to rotate about a clockwise, a counterclockwise direction.
01:53 So it's a positive moment, whereas our 100d is a negative moment.
01:58 And the distances we have are three meters, and then we have three meters across here, and and 4 meters across to point d.
02:15 And so if you look at the moment of b .y, we don't know what its force is.
02:20 We said it makes a positive moment, though, but the distance is 3 meters based on the diagram.
02:27 And the weight at point d was a negative moment.
02:30 Excerction to gravity 9 .81 meters per second, and the distance is 3 plus 4 or 7 meters, and this is all equal to 0.
02:40 So pretty easy algebra.
02:42 Move term over, divide by 3, and we see b y is 2 ,289 newton.
02:51 So now we can go back to sum of forces and see sum of forces in the y is equal to 0.
02:57 So we have a y, and we assume these unknowns in the positive direction just makes it simple.
03:01 If the answer is positive, it's in the positive direction.
03:03 If the answer is negative, it's in the negative...
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