ch06-p027

# ch06-p027 - 27. The free-body diagrams for the two blocks...

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For each block we take + x downhill (which is toward the lower-left in these diagrams) and + y in the direction of the normal force. Applying Newton’s second law to the x and y directions of both blocks A and B , we arrive at four equations: sin cos 0 sin cos 0 AA A NA A BB B NB B Wf T m a FW T m a θ −− = −= −+ = which, when combined with Eq. 6-2 ( Ak A N A f F μ = where k A = 0.10 and Bk B N B f F = f B where k B = 0.20), fully describe the dynamics of the system so long as the blocks have the same acceleration and T > 0. (a) From these equations, we find the acceleration to be 2 sin cos 3.5 m/s . kA A kB B AB WW ag μμ θθ §· + =− = ¨¸ + ©¹ (b) We solve the above equations for the tension and obtain () 0.21 N. T = + Simply returning the value for a found in part (a) into one of the above equations is certainly fine, and probably easier than solving for T algebraically as we have done, but
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