SOLUTION: 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

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

     =  =  = 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 =  =   and substitute it into equation (4)

     +  = 20

     = 20

     = 20

     = 20

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

    p             =  =  = 16 newtons.


ANSWER.  The magnitudes of tensity are  p= 16 N for CA  and  q= 12 N for CB.