(a) the horizontal force is 11.3 N
(b) The tension in the thread is 22.6 N
Draw a free body diagram for the suspended metal ball as shown in the figure. The tension T acts along the length of the thread, the force F is the horizontal force and its weight is mg, which acts vertically down wards.
Resolve the tension into two components- T cos 30 along the vertical and T sin30 along the horizontal as shown in the figure.
Apply the condition for equilibrium along the horizontal direction.
[tex]F=Tsin30........(1)[/tex]
Apply the condition for equilibrium along the vertical direction.
[tex]mg=Tcos30.......(2)[/tex]
Divide equation (1) by equation (2) and write an expression for F.
[tex]\frac{F}{mg} = tan30\\ F=mgtan30[/tex]
Substitute 2 kg for m and 9.81 m/s² for g.
[tex]F=mgtan30\\ =(2kg)(9.81m/s^2)(tan30^o)\\ =11.3N[/tex]
The horizontal force required to keep the ball taut is 11.3 N
From equation (2),
[tex]T=\frac{mg}{sin30} \\ =\frac{(2kg)(9.81m/s^2)}{sin30^o} \\ =22.6N[/tex]
Thus, the tension in the thread is 22.6 N
