Solutions To Physics, Engineering, And Chemistry Problems
Distance of the rocket from origin=150
Substituting and obtained in equations 1 and 2 into AB we obtain:
Optain all piont in this equations.
Torque by force about point A solve equation.
Torque of w about point A is
W= mg j
From equation 2,
Then, from equation 3
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.
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
2x +2y +4z =19000
2x +y +5z= 21000
We solve the equations using Gaussian elimination as follows:
(2x +y +5z) -(2x +2y +4z) =21000-19000
Subtracting equation 1 from equation 2 we obtain
Then, subtracting equation 3 from equation 2 we get
Subtracting equation 5 from equation 4 we obtain
Substituting into equation 5 we get
Substituting y and y into equation 1 we get
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.
When the position of the hypocycloid will be:
r(t) has an x-component only implying that the hypocycloid is a horizontal segment.
when r =r2
Hence, the acceleration is proportion to the distance oP