Question

QUESTION 2 A rear-wheeldrive, 2-Mg truck has an acceleration of 37 m/sec2, starting from rest. The coefficient of static friction between the wheels and the road surface is 0.8. The front wheels are free to roll. Neglect the mass of the wheels. Determine the normal force at Wheel A. G. 0.75 m 2 m — *-1.5 m Click Save and Submit to save and submit. Click Save All Answers to save all answers.

Answer 1

74000 N G 0.75 m 19620 N Y FfB .: .. M . NA NB 2 m-1.5 m-

Let the rosal seactions and fawetion Posice be denoted by NA, No and F and Fff respectively. The accelerating body can be transformed into a static body by adding an inertia force ie Firma , using D'Alembert's principle. Given a 37 m/s2 .: Firma = 2000 X 37 N = 74000 N This inertia force will act at the CG of the body in the direcion opposite to the direction of the acceleration, as shown. Also the weight of the body will au- vertically downwards at the CG as shown... Weight = mass X Acceleration due to gravity. = 2000 x 9.87 = 19620 N Now as the body is static, taking moment of all forces and equating their algebraic sum to zero.iie, EMA -0. Taking clochwise moment as positive and anticlockwise moment as negative. x 3.5 + Fax 0 + FRXO - 19620 X 1.5 - 74000 x 0.75 = 0 & NAX 3.5 + fx 0 + Fox & 3.5 NA = 84930 & MA= 24265.7 N the reaction or Normal force at Wheel A is 24265.7 N thence the reaction os 24.26 kN.