Question 1194515
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1a). A small block of mass 2kg is suspended by two strings of lengths 0.6m and 0.8m
from two points 1m apart on a horizontal beam. Find the tension in each string.
1b). A vehicle of mass 6 x 10³ kg moving with a velocity of 43kmh-¹ collides with a
stationary vehicle of mass 185 x 10³ kg and they stick together and move in the same
direction. They then came to rest after moving through a distance of 3m. Calculate the
magnitude of retardation.
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                    Problem 1a.



<pre>
Let A and B be the points on the horizontal beam where the two strings are attached.

Let C be the point where the mass is attached to the strings.


Notice that the triangle ABC is the right-angled triangle: its sides |AC| = 0.6 m
and |BC| = 0.8 m are the legs and its side |AB| = 1 m is the hypotenuse, since

    {{{sqrt(0.6^2 + 0.8^2)}}} = {{{sqrt(0.36 + 0.64)}}} = {{{sqrt(1)}}} = 1 meter.


Let vector P along CA be the tension of CA and let vector Q along CB be the tention of CB.

Let p and q be the magnitudes of these vectors, respectively.



Since the mass is in equilibrium,  the sum of x-components of vectors P and Q is zero:

    Px + Qx = 0,  or  p*cos(A) = q*cos(B).    (1)


Due to the same reason (Since the mass is in equilibrium), the sum of y-components of vectors P and Q 
is numerically equal to the weight of the mass m*g = 2*10 = 20 newtons (here for simplicity
I take g = 10 m/s^2 for the gravity constant).


It gives the second equation

    py + qy = 20 newtons,  or  p*sin(A) + q*cos(A) = 20.    (2)


Now notice that from triangle ABC,  sin(A) = 0.8/1 = 0.8;  sin(B) = 0.6/1 = 0.6;  cos(A) = 0.6/1 = 0.6;  cos(B) = 0.8/1 = 0.8.



Hence, you can write equations (1) and (2) in this equivalent form

    0.6p = 0.8q          (3)

    0.8p + 0.6q = 20     (4)


From equation (3), express p = {{{(0.8/0.6)*q}}} = {{{(8/6)q}}}  and substitute it into equation (4)

    {{{0.8*(8/6)q}}} + {{{0.6q}}} = 20

    {{{(6.4/6)q + 0.6q}}} = 20

    {{{(6.4/6)q + (3.6/6)q}}} = 20

    {{{(10/6)q}}} = 20

    q = (6*20)/10 = 12 newtons.

    p             = {{{(8/6)*q}}} = {{{(8/6)*12}}} = 16 newtons.


<U>ANSWER</U>.  The magnitudes of tensity are  p= 16 N for CA  and  q= 12 N for CB.
</pre>

Solved.



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Regarding your second problem, &nbsp;it is posed incorrectly, &nbsp;since it uses the term &nbsp;" retardation ",
which is not appropriate, &nbsp;not applicable and is &nbsp;IRRELEVANT &nbsp;in this context.


The appropriate terms are &nbsp;EITHER &nbsp;the &nbsp;" deceleration " &nbsp;OR &nbsp;the  &nbsp;" friction force ".



NEVER &nbsp;SUBMIT &nbsp;more than one problem per post.



It is the &nbsp;RULE &nbsp;of this forum.



About the rules, &nbsp;read on web-page


https://www.algebra.com/tutors/students/ask.mpl?action=ask_question&topic=Equations&return_url=http://www.algebra.com/algebra/homework/equations/


from which you submit your problems.