The acceleration of a 3.00 kg object under three forces is found to be a vector

Question

The acceleration of a 3.00 kg object under three forces is found to be a vector = (-12.00i^+ 9.00j^)m/s^2. If two of the forces are F_1 vector = (30i^+ 16j^+ 2k^)N and F_2 vector = (-12i^+ 8j^- 5k^)N, find the third force. A slab of mass M rests on a frictionless floor. A block of mass m rests on the top of the slab. The coefficients of statics and kinetic frictions between the block and slab are, respectively, mu_s and mu_k. The block is pulled by a horizontal force F as shown in Fig. 5. (a) Draw the free body diagrams for block and slab. (b) Apply Newton's second law and determine the direction and magnitude of the acceleration of block. (c) Determine the direction and magnitude of acceleration of the slab.

Explanation / Answer

11. from newton's 2nd law,

Fnet = m a

F = 3(-12i + 9j) = -36i + 27j N

and F1 + F2 + F3 = F

(30i + 16j + 2k) + (-12i + 8j - 5k) + f3 = -36i + 27j

F3 = - 54i + 3j + 3k .......Ans

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