a.
Distance of the rocket from origin=150
b.
c.
2.
a.
but
So,
Substituting and obtained in equations 1 and 2 into AB we obtain:
Optain all piont in this equations.
b.
Force
Torque by force about point A solve equation.
c.
Torque of w about point A is
W= mg j
From equation 2,
Then, from equation 3
d.
The resultant torque about point equation
The equation has two z components. That is, and
For the resultant torque to be negative, the latter component should be larger than the first component. That is,
f(2r-h)cos 0> f (sin0+ mg) 2 hr – h sqaure and root.
e.
We have:
Substituting the given variables into the inequality we get:
Dividing both sides by we obtain
The heaviest mass that can be pulled 73.215 kg
3.
2x +2y +4z =19000
2x +y +5z= 21000
2x+3y+4z=20000
We solve the equations using Gaussian elimination as follows:
(2x +y +5z) -(2x +2y +4z) =21000-19000
-y+z=2000
Subtracting equation 1 from equation 2 we obtain
-2y+z=2000
Then, subtracting equation 3 from equation 2 we get
2000-1000= 1000
y =1000
Subtracting equation 5 from equation 4 we obtain
Substituting into equation 5 we get
Substituting y and y into equation 1 we get
Therefore
b(i)
b(ii)
batch 1 total cost 19000
batch size 10 per cost batch 19250/5 =18250
batch size 10 per cost batch 183000/10=18300
The most cost effective batch size is 5 since it has the lowest unit cost.
4.
a.
When the position of the hypocycloid will be:
r(t) has an x-component only implying that the hypocycloid is a horizontal segment.
b.
Given that
c.
when r =r2
acce r”(t)
Hence, the acceleration is proportion to the distance oP
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