Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Answer: 0.169 (3 s.f.)
Explanation:
Force = 76 N
Spring constant = 450 N/m
Extension/displacement = x
Hooke's law states that: F = kx
Therefore, 76 = 450 X x
76/450 = x
0.169 (3 s.f.) = x
Answer:
Energy
Kinetic
Energy in
this
Explanation:
ithikitsthatecauseireallydo
Answer:
a) {[1.25 1.5 1.75 2.5 2.75]
[35 30 25 20 15] }
b) {[1.5 2 40]
[1.75 3 35]
[2.25 2 25]
[2.75 4 15]}
Explanation:
Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]
[1 2 3 1 2 3 4]
[45 40 35 30 25 20 15]}
Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.
Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).
For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.
For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:
{[1.5]
[2]
[40]}
Transposing this column will make our first row of K look like this:
{[1.5 2 40]}
Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.
