The resultant of two forces P and Q is R. if ofne of the forces be reversed in direction , the resutant is S show that R^2+s^2= 2 (P^2+Q^2).

Two forces act an angle 120 degree. The greater is represented by 80kg wt. and the resultant is at right angle to lesser .Find the latter.

Find the angle between two equal forces P, P when the squere of the resultant is equal to( \( {-2+ \sqrt{3} }\)) times their product.

The resultant of two forces P and Q aaaaact at right angle to P. show that the angle between the forces is cos^(-1)(-P/Q). also show that R^2= Q^2-P^2.

Two forces acting at a point are such thst if the direction of one is reveresed , the direction of resultant is turned through right angle . Prove that the forse must be equal in magnitude.

Two forces P and Qacting along two intersecting lines ,having resultant R, If the force Q is reversed in direction  and changed in to (R^2-P^2)/Q , show that the resultant is still of magnititude R.

The resultant of two forces P and Q acting at angle \(\theta\) is equal to (2m+1)\( \sqrt{P^2+Q^2} \)and when they acti at angle \(90^{\circ}\)-\(\theta\),the resultant is (2m- 1)\( \sqrt{P^2+Q^2} \). prove that \(tan\theta ={m-1}/{m+1}\)

The resultant of two forces P and Q acting along the line OA and OB respectively is at right angle toOA and the resultant of forses P' abd Q' actin respectively along the same straight line is right angles toOB. Show that PP'=QQ'.

The resultant  of two forces P and Q is perpendicular to P ; the resultant of P and Q' acting at same angle is perpendicular toQ' . prove that P is the geometric mean between Q and Q'.

Two forces P and Q have a resultant R. If the forces P be increased , then the new resultant bisects the angle between R and P . Find the increase in P.

The resultant R of the force P and Q makes an angle \(2\theta\) with the line of acting of P. P is now replaced by P+R , Q remaining unchanged . Show that the resultant makes an angle \(\theta\) with  P.