Question

Problem 15.125

Water is discharged from a nozzle with a velocity of 8 m/s and strikes the blade mounted on the 18-kg cart. The nozzle has a diameter of 50 mm and the density of water is ρw = 1000 kg/mg3.

A: Determine the tension developed in the cord, needed to hold the cart stationary.

B: Determine the normal reaction of the wheels on the cart.

Answer 1

Answer: Page 1 15-125 ) praw the free body diagram of the control volume. flow rate of the water by calculate the mass Using relation at - Swater Q = Swater Av Swater I du Here, swater is the density of the water, d is the diameter of the nozzle, and vis the velocity of the water discharge from the nozzle.

page ② Substitute 1000 kg/m² for Swater, 0.05 m ford, and 8 m/s for v am de = 1000x T X0.05) 'x8 = 1000 X 0.7853X0.0025 X 8 = 15.706 kg/s calculate the -component of the force of the oblade . -Fx - do Cave) -Fe = dm ((Ve), -(\)) substitute 15.706 Igls for dm ucos us for (VB) and & for (VA) a dt then, F = 15.706 (w cosusu) = 15.706v(cos4S-1) substitute 8 m/s for v. cs Scanned with

page (² -F = 15-706X8( cosuso-1) = 15.706 X 8 (0-7071-1) = 15.706 X8 (-0.2929) Fx = -36.8N Fx = 36.8N force of calculate the y-component of the the cable fy = am (Avy) de • de Substitute 15.706 kg ls for dm usinus for (VB) y and o for (VA) y . Fy = 15.706 (v sinus-o) = 15.706 V sin usº substitute 8m/s for v Fy = 15.706 X 8 Sin 45° = 15.706 X 5-6568 = 88.84 N

page 4 Draw the free body diagram of the carl- calculate the tension in the cord, and normal reaction of the wheels by using the equilibrium equations, F = 0 : Fx - T=0 substitute 36.8N for Fx 36.8-T=0 T=368N Therefore, the tension in the cord is 368 N

Pages { Fy=0 N-mcg - Fy=0 flere me is the mass of the cart, and g is the acceleration due to gravity. substitute 18kg for me, 88.84N for Fy, and 9.81 mls for g.. N- 18X9,81-88.84 = 0 N-176.58 - 8884 = 0 N- 265.42=0 N = 265.42 N of the Therefore, the normal reaction wheets is 265.42 N