Problem 2P

A jet plane traveling 1890 km/h (525 m/s) pulls out of a dive by moving in an arc of radius 6.00 km. What is the plane’s acceleration in g’s?

Solution: To calculate magnitude and direction of third rope,we can use newton’s second law it states the total net force is equal to mass and acceleration of the body. F = ma (m) is mass of the body and (a) is acceleration of the body. Below figure shows the free body diagram of the rope system. In The above figure shows the force in rope 1 and rope 2 must balance the force in rope 3.rope 3 is an angle needs to break the x and y-directions.is the angle of third rope with the horizontal,T 3osis the horizontal component of the tension in third rope in positive x-direction and T sin is the vertical 3 component of the tension in third rope in negative y-direction. The net force acting on horizontal direction is zero. F =x0 2 (T3cos) T 1 0 Then T 3os = T 1 1 The net force acting on vertical direction is zero F x 0 2 (T3sin) T 2 0 Then T 3in = T 2 2 Square and add equations 1 and 2 2 2 2 2 (T 3os) +(T si3) =T 1 +T 2 2 2 2 2 2 T 3cos +sin )=T 1 +T 2 T 2= T 2 + T 2 3 1 2 Then substitute T =10 N and T =80 N 2 T 2= (50 N) + (80 N) 2 3 T 3 94.3 N Therefore magnitude of the tension in third rope is T3= 94.3 N Divide equation 2÷equation 1 T3sin/T 3os =T /2 1 tan = T 2T 1 1 =tan ( 2 /T1) Then substitute T 150 N and T =80 N2 1 =tan ( 80/50) =57.99 ~58 0 0 The tension in the third rope is directed 58 below the horizontal .