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To satisfy T ti f design d i limitations li it ti it is i necessary to determine the effect of the 2-kN tension in the cable on the shear, tension, and bending of the fixed I-beam. I beam For this purpose replace this force by its equivalent of two forces at A, Ft parallel beam and Fn perpendicular to the beam. Determine Ft and Fn.

Ft = 2cos50D Ft = 1.28 kN Fn = 2sin 50D Fn = 1.53 1 53 kN

Determine D i the h magnitude i d Fs off the tensile spring force in order that the resultant of Fs and F is a vertical force. Determine the magnitude R of this vertical g resultant force.

Fs cos 60 = 120 Fs = 60 lb R sin 60D = 120 R = 103.92 lb D

IIn the h design d i off a controll mechanism it is determined that mechanism, rod AB transmits a 260 N force P to the crank BC. Determine the x and y scalar components of p P.

5 12 -1 ⎛ 5 ⎞ θ = tan ⎜ ⎟ ⎝ 12 ⎠ θ = 22.62 22 62D Px = −260cos ( 22.62D ) Px = −240 N Py = −260sin ( 22.62D ) Py = −100 100 N tan θ =

For the h mechanism h i shown, h d t determine i the th scalar l components Pt and Pn of P which are tangent and normal, normal respectively, to crank BC.

tan α =

5 12D

α = 22.62

β = 30 − α = 30D − 22.62D D

β = 7.38D Pn = 260cos ( 7.38D ) Pn = 257.84 N Pt = 260sin ( 7.38D ) Pt = 33.39 33 39 N

Iff the h equall tensions i T in i the h pulley ll cable bl are 400 N, N express in vector notation the force R exerted on the pulley by the two tensions. Determine the g of R. magnitude

Rx Rx Rx Ry Ry Ry

= ∑ Fx = 400 + 400cos 60D = 600 N = ∑ Fy = 400sin 60D G G G = 346.41 N R = 600i + 346.41 j R = ( 600 ) + ( 346.41) R = 692.82 692 82 N 2

2

While steadily pushing the machine up an incline, incline a person exerts a 180 N force P as shown,, Determine the components of P which are parallel and perpendicular to the incline.

Pt = 180cos 25D Pt = 163.13 N

Pn = −180sin 25D Pn = −76.07 N

The normal reaction force N and the tangential friction force F act on the tire of a front wheel drive car as shown. Express the resultant R of these two forces in terms of the unit vectors ((a)) i and j along the x-y axes and (b) et and en along the n-t axes shown.

G (a ) Rx = [ 400cos15D + 900cos105D ] iˆ lb G Rx = 153.43 153 43iˆ lb G Ry = [ 400sin15D + 900 sin105D ] ˆj lb G Ry = 972.86 ˆj lb

G R = 153.43iˆ + 972.86 ˆj lb G (b) R = 400eˆt + 900eˆn lb

Determine D t i the th resultant lt t R off the th two t forces f applied li d to the h bracket. b k Write W i R in i terms off unit i vectors along the x-and y-axes.

Rx = ∑ Fx Rx = 200cos35D + 150cos120D Rx = 88.83 88 83 N Ry = ∑ Fy Ry = 200sin 35D + 150sin120D Ry = 244.61 N G G G R = 88.83i + 244.61 j N

The ratio of the lift force L to the g force D for the simple p airfoil is drag L/D=10. If the lift force on a short section i off the h airfoil i f il is i 50 0 lb lb, compute the magnitude of the resultant force R and the angle θ which it makes with the horizontal.

L = 10, L=50 lb D 50 D = = 5 lb 10 R = L2 + D 2 R = ((50)) 2 + ((5)) 2 = 50.25 lb −1 ⎛ L ⎞ −1 ⎛ 50 ⎞ θ = tan ⎜ ⎟ = tan ⎜ ⎟ ⎝ D⎠ ⎝5⎠ θ = 84.28 84 28D