Two workers are moving a slab of ice (m = 290 kg) across the factory floor. As seen from above, one worker exerts a force of 415 N at an angle of -20o, and the other exerts a force of 470 N at an angle of 50o. What is the magnitude of the acceleration of the ice slab?

Answer 1

Fr = 415[-20o] + 470[50o].
Fr = (415*Cos(-20)+470*Cos50) +
(415*sin(-20)+470*sin50).
Fr = (390+302) + (-142+360)i.
Fr = 692 + 218i = 726N[17.5o].

Fr = M*a; a = Fr/M = 726[17.5]/290 = 2.50m/s^2[17.5o].