Find the slope and y-intercept of the line whose equation is given.
1 answer:
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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)
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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
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The sum of forces is ...
F = a +b +c +d = (92.337, -17.314) = 93.946∠-10.62° . . . N
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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.
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<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.
Answer:
Step-by-step explanation:
You know how subtraction is the <em>opposite of addition </em>and division is the <em>opposite of multiplication</em>? A logarithm is the <em>opposite of an exponent</em>. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.
We know that ; as a logarithm, the <em>exponent</em> gets moved to its own side of the equation, and we write the equation like this:
, which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."
One historical quirk: because we use the decimal system, it's assumed that an expression like uses <em>base 10</em>, and you'd interpret it as "What power do I raise 10 to to get 1000?"
The expression means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is
.
(If we wanted to, we could also solve this: , so
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