Truck On a Hill

Section 6.4

This problem ends Section 6.4 for us. I need a step by step solution for all three parts. Thank you.

 

(a) Find the force required to keep the truck from rolling down the hill in terms of d

because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector
F=-30000j

to find the forse required to keep the truck from rolling down the hill, project F onto a unit vector v in the direction up hill, as follows:

v=cos(d°)i+sin(d°)j

therefore, projection of F onto v is vector w1

w1 = (F*v)/||v^2||*v.................since F=-30000j and ||v^2||=sqrt(cos^2(d)+sin^2(d))=1, we have
w1 = (-30000j*(cos(d°)i+sin(d°)j))/|*(cos(d°)i+sin(d°)j)
w1=(-30000 i*j (cos(x) -30000sin(x) j^2)*(cos(d°)i+sin(d°)j).......ij=0,j^2=1
w1=(-30000*0 (cos(x) -30000sin(x) *1)*(cos(d°)i+sin(d°)j)
w1=( -30000sin(x) (cos(d°)i+sin(d°)j)
w1= - 30000sin(d) cos(d)i -30000 sin^2(d)j



then:
w2 =F-w1
w2 =-30000j-(- 30000sin(d) cos(d)i -30000 sin^2(d)j)
w2 =-30000j+ 30000sin(d) cos(d)i +30000 sin^2(d)j
w2 =30000sin(d) cos(d)i +30000 sin^2(d)j-30000j
w2 =30000sin(d) cos(d)i +30000( sin^2(d)-1)j


b.
use w2 to fill out the table

w2=30000sin(1°) cos(1°)i +30000( sin^2(1°)-1)j
=30000*0.0175i+30000*(-0.9997)j
=525i-29991j
||w2||=force=sqrt(525^2+(-29991)^2)=29995.6

so,
.......d|1°....................|2°............3°
force|29995.6..........|................|

you can continue with d=2°,d=3°,.....

c.

w2=30000 sin(5°) (cos(5°) i) + 30000 (sin(5°)^2 - 1) j

w2=30000 (0.08715574274765817) (0.9961946980917455i) + 30000 ((0.08715574274765817^2 - 1) j

w2=2604.723i -29772.116 j

||w2||=sqrt(2604.723^2+(-29772.116)^2)=29885.8 pounds

 

This is precisely what I was looking for. Thanks.

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