Problem 4.118 Part A Replace the loading acting on the beam by a single resultan

Question

Problem 4.118 Part A Replace the loading acting on the beam by a single resultant force Egure.1) Determine the magnitude of the resultant force, it Fi 460 N F 330 N, and F 710 N Express your answer to three significant figures and include the appropriate units F: 1350 N Submit My Answers Give Up Correct Part B Determine the angle between the resultant force and the z axos Express your answer to three significant figures and include the appropriate units. Figure 1 ot : 84 7, courted counterclockwise torn negative z axis Submit My. Answers Give Up Correct Part C Specitly where the force acts, measured from B Express your answer to three significant figures and include the appropriate units 7.23 1m Type here to search

Explanation / Answer

Given F1 = 460 N, F2 = 330 N, F3 = 710 N

A. let i and j be unit vectors in x and y direction

then From the figure

F1 = 460(cos(60)i - sin(60)j)

F2 = -330j

F3 = 710(-sin(30)i - cos(30)j)

so resultant

F = F1 + F2 + F3 = -125i - 1343.249 N

Magnitude of resultant

|F| = sqroot(Fx^2 + Fy^2) = 1349.05 N

B. angle between resultant force and x axis = theta

then tan(theta) = 1343.249/125

theta = 84.63 deg

C. consider a point to the left of point B by x m

moment at this point should bve zero due to the forces

then , F1sin(60)*(4 - x) = F2*x + F3cos(30)*(3 + x)

398.37(4 - x) = 330x + 614.87(3 + x)

330x + 614.87x + 398.37x = 398.37*4 - 3*614.87

x = -0.1869

so the force acts at x = 0.1869 m to the right of point B

Q2. Let i, j and k be unit vectors in x, y and z directions

then, F1 = -300 k

F2 = 200 j

F3 = -250i + 350j - 450 k

then resultant force, F = F1 + F2 + F3 = -250i + 550 j - 750 k

moment about O ->

Mo = 300*2i + 200*1.5k + (1.5 i + 2j) x (-250i + 350j - 450 k)

Mo = 600i + 300k + 525k + 675j + 500k - 900i = -300i + 675j + 1325k Nm

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