A2.0kg disk is accelerated at 8.0m/s2 across a frictionless surface by three hor

Question

A2.0kg disk is accelerated at 8.0m/s2 across a frictionless surface by three horizontal forces.Forces F1and F2have magnitudes 10.0N and 12.0N, respectively.The figure is an incomplete diagram for the situation, with theacceleration a included. Whatis the third force F3? Sorry its reallysloppy, the custom diagram wouldn't add the first detailed one idid. The angle between force 1 and the negative x axis is 60degrees, and the angle between the acceleration and the negative yaxis is 30 degrees. I would appreciate ANY help you may offer.Thank you very much.

Explanation / Answer

. 1.   I shall use bold face type to denotevectors. 2.   I shall use "âx" and"ây" to denote the unit vectors in thex-direction and y-direction, respectively. . 3.   F4 = force of the 2.0kg disc at 8.0m/s2 = (mass)*(acceleration) = (2.0kg)*(8.0m/s2) 4.   F4 = 16.0 Newtons @300 degrees = 8.0âx -13.86ây 5.   F1 = (10Newtons)*[cos(240 degerees)]âx + (10 Newtons)*[sin(240degerees)]ây 6.   F1 =-5.0âx-8.66ây 7.   F2 =0.0âx +12.0ây 8.   Given the above for F4, F2, and F1, we cannow calculate F3: 09.   F3 = F4- F2 - F1 10.   F3 = [8.0 - 0.0 -(-5.0)]âx + [-13.86 -(-8.66) -(12.0)]ây 11.   F3 = 13.0 Newtonsâx - 17.2 Newtonsây 12.   F3 = 21.56 Newtons @307.08 degrees .

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