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o sa s y es gn m a ons s necessary

to determine the effect of the 2-kNtension in the cable on the shear, tension,

- .

purpose replace this force by its

equivalent of two forces at A, Ft parallel

n .Determine Ft and Fn.

t=

1.28 kNt

F=2sin50

nF =

.n=

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Determ ne t e magn tu e Fs

o

e ens e spr ng orce n or er

s

ma nitude R of this verticalresultant force.

cos120

s=

60sF =

sin60120

=

103.92 lbR=

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In t e es gn o a contro

mec an sm, s e erm ne a

thex and scalar com onents of P.

5tan=

-1 5

12

260cos 22.62P = .=

240 Nx

P =

260sin 22.62yP =

y

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For t e mec an sm s own,

e erm ne e sca ar

t n

,

, .

5tan

12=

22.62=

30 30 22.62 = =

260cos 7.38P = 7.38=

n

257.84 Nn

P =

( )260sin 7.38tP=

.t

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I t e equa tens ons T n t e

pu ey ca e are , express

.ma nitude of R.

x xR F=

400 400cos60x

R = +

x =R F=

y y

400sin60y

R =

346.41 Ny

R = 600 346.41R i j= +

( ) ( )2 2

600 346.41R= +.=

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,exerts a 180 N force P as shown

Determine the components of P

which are parallel and

perpendicular to the incline.

180cos25t

P=

163.13 NtP=

180sin25P =

76.07 Nn

P =

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The normal reaction forceNand the

tangential friction force Fact on the tireof a front wheel drive car as shown.

forces in terms of the unit vectors a i

and j along thex-y axes and (b)etand e

nalong the n-t axes shown.

( ) 400cos15 900cos105 lbx

a R i= +

400sin15 900 105 lbR sin j= +

.x

972.86 lbR j=

153.43 972.86 lbR i j= +

( ) 400 900 lb t nb R e e= +

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e erm ne e resu an

o e wo orces app e

to t e rac et. r te n

terms of unit vectors

along thex-andy-axes.

x xR F=

200cos35 150cos120x

R = +

.x=R F=

y y

200sin35 150sin120y

R = +

244.61 Ny

R =

88.83 244.61 NR i j= +

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drag forceD for the simple airfoil isL/D=10. If the lift force on a short

sect on o t e a r o s 50 ,

resultant forceR and the an le which it makes with the horizontal.

10, L=50 lbL

=

50

102 2

2 250 5 50.25 lbR= + =

1 1 50tan tanL

= =

5D= .