SOLUTION: Hello I have a few problems that I can't get, if anyone could help me, that would be great! 1) A sliding door moves along the line of vector PQ. If a force is applied to the doo

Algebra ->  Trigonometry-basics -> SOLUTION: Hello I have a few problems that I can't get, if anyone could help me, that would be great! 1) A sliding door moves along the line of vector PQ. If a force is applied to the doo      Log On


   

Question 118679: Hello I have a few problems that I can't get, if anyone could help me, that would be great!
1) A sliding door moves along the line of vector PQ. If a force is applied to the door along a vector that is orthogonal to PQ then no work is done. True or False, justify your answer.
Orthogonal is perpendicular, right? So if force is applied to the long side of the door, then the door moves so work is done? But how can you show this mathematically?
2) sin2x - sqrt3sinx = 0 FInd exact solutions of the equation in the interval [0,2PI)
I got to as far as 2sinx-cosx - sqrt3sinx = 0
Thank you in advance to any tutors that help!
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Your problem:
.
The answer to the first question is true ... no work is done. The definition of work is based
on the force applied times the distance that the object moves IN THE DIRECTION OF THE APPLIED
FORCE. But in this case, there is no movement in the direction of the orthogonal force. The
door moves perpendicular to this orthogonal force, so there is no movement in the direction
of the force. If the door is moving, its movement is caused by another force, but not by the
orthogonal force.
.
The general form of the equation for work is:
.
Work = Force * Distance * cos A
.
where A is the angle between the applied force and the direction of movement. And in this case
the Force is at 90 degrees to the distance being moved. Since cos 90 degrees is zero, the
work done is:
.
Work = Force * Distance * cos 90 = Force * Distance * 0 = 0
.
Second problem:
.
sin%282x%29+-+sqrt%283%29%2Asinx+=+0
.
Applying the double angle formula to the first term changes the equation to:
.
2%2Asin%28x%29%2Acos%28x%29+-+sqrt%283%29%2Asinx+=+0
.
Notice that sin(x) is a common factor on the left side. Factoring sin(x) from this equation
changes it to:
.
sin%28x%29%282%2Acos%28x%29+-+sqrt%283%29%29+=+0
.
This equation will be true if either of the factors on the left side is equal to zero because
a multiplication by zero on the left side will make the entire left side equal to zero and
therefore equal to the right side.
.
So first, what values of x will make sin(x) equal to zero? The sin function is equal to
zero at 0 degrees (0 radians) and at 180 degrees (pi radians) and again at 360 degrees
(2 pi radians).
.
Then look at the second factor %282%2Acos%28x%29+-+sqrt%283%29%29 and setting it equal to zero results
in the equation:
.
2%2Acos%28x%29+-+sqrt%283%29+=+0
.
Get rid of the -sqrt%283%29 on the left side by adding sqrt%283%29 to both sides to make
the equation:
.
2%2Acos%28x%29+=+sqrt%283%29
.
Then divide both sides by 2 and get:
.
cos%28x%29+=+sqrt%283%29%2F2+
.
Taking the arccos to solve for x returns the answer that x+=+pi%2F6 which is x = 30 degrees
and also x+=+-pi%2F6+=+%2811%2Api%29%2F6 which is x = 330 degrees.
.
So there are 5 solutions over the interval 0 degrees through 360 degrees. Those solutions are:
.
zero degrees (or 0 radians)
180 degrees (or pi radians)
360 degrees (or 2%2Api radians)
30 degrees (or pi%2F6 radians)
330 degrees (or %2811%2Api%29%2F6 radians)
.
Hope this helps you with these two problems and shows you how to solve them.