a)The tension in each cable forms a triangle as shown below
The triangle formed is a right triangle.
AC represents the inclined cable
AB represents the horizontal cable
BC represents the vertical cable
The tension is the force acting on the vertical table. Given that the force is 512N,
the tension on the vertical table is 512N
To find the tension on the inclined cable, we would apply the Sine trigonometric ratio which is expressed as
Sinθ = opposite side/hypotenuse
From the triangle,
θ = 33
opposite side = BC = 512
hypotenuse = AC
Thus,
Sin 33 = 512/AC
By crossmultiplying,
512 = ACSin33
AC = 512/Sin33
AC = 940.1 N
The tension on the inclined cable is 940.1 N
To find the tension on the horizontal cable, we would apply the tangent trigonometric ratio which is expressed as
Tanθ = opposite side/adjacent side
adjacent side = AB
tan 33 = 512/AB
By crossmultiplying,
ABtan33 = 512
Dividing both sides of the equation by tan 33, we have
AB = 512/tan33
AB = 788.4
The tension on the horizontal cable is N788.4
