A ball weighing 50N is pulled back by a rope to an angle of 20 degreees. What is the tension in the pulling rope?

Question

 Hi everyone. I really dont know how to solve this. I sttempted to make a coordinate graph and randonly started solving a bunch of sides of the angle but this is wrong. I dont know how to do this problem. The problem comes with an image of the situation http://i1076.photobucket.com/albums/w457/Hbjoi_Hggo/image.jpg1_zpsrevntim4.jpg

Answer 1

It looks like they want you to find the tension in the horizontal rope pulling to the left.  Since it is perpendicular to the weight, it would have to balance the horizontal component of the the tension in the rope that is inclined 20° from the vertical.  Let us start with that one.  Call it T₁.  The vertical component of T₁ must balance the 50 N weight.

 

T₁cos20° = 50 N ⇒ T₁ = (50 N) / (cos20°)

 

Call T₂ the tension in the rope pulling to the left.  It must balance the component of T₁ pulling to the right.

 

T₂ = T₁sin20° = [(50 N) / (cos20°)] sin20° = (50 N) tan20° = 18.2 N

 

That is the tension in the rope pulling to the left.

Answer 2

This is a statics problem.  The drawing shows you how to construct the free body diagram.  When you pull the ball to the left 20 degrees, the force acting on the ball is split into two components.  Think of a right triangle.  The 50N is the force pulling vertically down on the ball.  The small angle at the top is 20 degrees.  They are asking you to find the hypotenuse.  So cos(20) = 50N/answer.   Answer = 50 N/cos(20)