The pass mark for a test is 86%. Steve scores 52 out of 61 marks. Does he pass the test?
2 answers:
You might be interested in
9514 1404 393
Answer:
- resultant force: 93.946∠-10.62° N
- line of action: 17.314x +92.337y = 809.433
Step-by-step explanation:
We can use the notation a∠b to represent the (x, y) components (a·cos(b), a·sin(b)), where angle b is measured CCW from the +x direction. If we label the forces a, b, c, d clockwise from A, then we have ...
a = 80∠0° = (80, 0)
b = 60∠90° = (0, 60)
c = 90∠45° = (63.640, 63.640)
d = 150∠-110° = (-51.303, -140.954)
__
If we label point A the origin, then the clockwise torque on point A is the sum of products of the force x-component and its y location, and its y-component and the negative of its x location.
T = (0, 0)·(80, 0) +(3, 0)·(0, 60) +(3, -8)·(63.640, 63.640) +(0, -8)·(-51.303, -140.954)
T = 809.433 . . . . n·m, the CW torque on point A
__
The sum of forces is ...
F = a +b +c +d = (92.337, -17.314) = 93.946∠-10.62° . . . N
__
This force, applied to the point of application, must generate the same torque as the given forces. That is ...
F·(y, -x) = 809.433
Then the equation of the line of action is ...
17.314x +92.337y = 809.433 . . . . . x and y in meters measured from A
Any point (x, y) on this line will serve as a point of application of the force. Unfortunately, this line of action does not pass through the rectangular plate. The attachment shows the point (D) on the line of action that is closest to point A.
_____
<em>Additional comment</em>
The resultant force could be decomposed into two forces acting <em>on the rectangular plate</em>. One could be of much larger magnitude, operating at the corner opposite point A. This force would provide the necessary torque. Another would be acting on point A, providing no torque, but with components such that the resultant has the correct magnitude and direction.
