Question #03abb

2 Answers
Dec 31, 2015

Answer:

#/_ PAQ =60^o#

Explanation:

Tony B

The trick is to nearly always draw a diagram and this will help the understanding of what is going on.

Given that: #color(white)(..)tan^(-1)(sqrt(3)/2) = theta#
and that PC = 2 units marries up very well with triangle PBC. This shows that this diagram is correct.

PC is parallel to AQ so

#color(blue)("Force diagram PAQC reviewed against question and confirmed")#

#/_BAQ = /_BPC = x+theta->color(blue)("Checked and confirmed")#

#/_ theta =tan^(-1)(sqrt(3)/2)-> color(blue)(" As given and confirmed")#
Thus #BC=sqrt(3)" and "BA=2#

#color(magenta)("PA=1 unit from ratio of 2:1") =>PB=1 larr" Updated"#

#color(blue)("Thus BPAC structure confirmed")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Pythagoras and the given of:
#BC=sqrt(3)color(white)(..) ,color(white)(..) PC = 2#

#BP = sqrt( 2^2 - (sqrt(3))^2) = 1 color(blue)(->" BP confirmed")#

#color(blue)(Delta" BPC structure confirmed")#

#color(red)("Whole diagram is confirmed as correct.")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus #/_ (x+theta)^o = /_BPC ->color(blue)(" Confirmed")#

We are asked to find: #/_ (x+theta)^o#

From the diagram: #/_ PAQ = (x+theta)^o = tan^(-1)(sqrt(3)/1)#

#/_ PAQ =60^o#

Jan 1, 2016

Answer:

The answer is (4) #60^@#

Explanation:

The idea here is that you need to draw the #x#-projection and the #y#-projection of the resultant, and use the given #1:2# ratio that exists between #P# and #Q# to find the tangent of the given angle.

Now, you are told that an angle of

#color(red)(alpha) = tan^(-1)(sqrt(3)/2)#

exists between the resultant, which we'll call #color(blue)(R)#, and #P#. This means that the tangent of this angle will be

#tan(color(red)(alpha)) = [tan^(1-)(sqrt(3)/2)] = sqrt(3)/2#

Keep this in mind. Now, here's a very rough sketch of the two forces

enter image source here

So, #P# is drawn on the #x#-axis and the angle #color(red)(alpha)# is shown in #color(red)("red")# here.

The resultant, #color(blue)(R)#, will have two projections, one on the #x#-axis, #R_x#, and the other on the #y#-axis, #R_y#.

The idea here is that you need to write #R_x# and #R_y# in terms of #P#, #Q#, and #color(green)(beta)#, the angle that exists between the two forces - shown here in #color(green)("green")#.

Now, focus on the triangle marked with light blue lines. Using the angle #color(green)(beta)#, you can say that

#R_y = Q * sin(color(green)(beta))#

Here #Q# is the hypotenuse of the aforementioned triangle.

At the same time, you can say that the leg of this triangle located on the #x#-axis will be equal to #Q * cos(color(green)(beta))#, which means that #R_x# will be equal to

#R_x = P + Q * cos(color(green)(beta))#

Use the given #1:2# ratio that exists between #Q# and #P# to write

#{(R_y = 2P * sin(color(green)(beta))), (R_y = P + 2P * cos(color(green)(beta))) :}#

Finally, focus on the triangle formed by the two projections of the resultant ans the resultant. You know that

#tan(color(red)(alpha)) = sqrt(3)/2#

At the same time,

#tan(color(red)(alpha)) = R_y/R_x#

This means that you have

#sqrt(3)/2 = (2 color(red)(cancel(color(black)(P))) * sin(color(green)(beta)))/(color(red)(cancel(color(black)(P))) + 2color(red)(cancel(color(black)(P))) * cos(color(green)(beta)))#

#sqrt(3)/2 = (2sin(color(green)(beta)))/(1 + 2cos(color(green)(beta)))#

This is equivalent to

#sqrt(3) + 2sqrt(3) * cos(color(green)(beta)) = 4 * sin(color(green)(beta))#

Looking at the options given to you, the only one that matches this equation corresponds to #color(green)(beta) = 60^@#, since

#sqrt(3) + 2sqrt(3) * cos(60^@) = 4 * sin(60^@)#

#sqrt(3) + color(red)(cancel(color(black)(2)))sqrt(3) * 1/color(red)(cancel(color(black)(2))) = 4 * sqrt(3)/2#

#2sqrt(3) = 2sqrt(3) color(white)(x)color(green)(sqrt())#