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xzibition8612
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Homework Statement
The 200-lb block is at rest on the floor (u=0.1) before the 50-lb force is applied as shown in Figure 1. What is the acceleration of the block immediately after application of force? Assume the block is wide enough that it cannot tip over.
Homework Equations
F=ma
The Attempt at a Solution
See my free body diagram (FBD attachment).
I setup the equations:
-fi+Fcos30i+Nj-mgj-Fsin30j=m(x''i+y''j+z''k) (i,j,k are unit vectors, x'',y'',z'' are double dot or acceleration)
Thus I get two equations:
Fcos30-f=mx''
mg+Fsin30=N
I plug in for the second equation and get N=6425. Coefficient of static friction is 0.1, so my max static friction is 0.1*6425 = 642.5. So if I go back to the first equation, it looks like the block doesn't move despite the 50lb force. But the book gives the answer as 3.55ft/s^2 in the positive x direction, so I screwed up somewhere. But I can't see it. Any help would be appreciated.