- #1
freecorp777
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Here's a pulley problem that has got me stopped; I'd appreciate any help that's offered.
There are two blocks, A (44 N) and B (22 N) connected via a rope that stretches over a pulley. Block A is resting on a table, and block B is hanging over the edge. Block C is positioned on top of Block A.
The question reads: "Given that the coefficient of static friction between Block A and the tabel is 0.20, what is the minimum weight of Block C to keep Block A from sliding off the table?"(The answer that is given is 66 N, but I can't seem to get this)I tried starting off with:
Block A=M=44N Block B=m=22N
(let F stand for the friction force)
T-F=Ma and mg-T=ma
Solving for Tension and setting them equal gives:
mg-ma=F+Ma
Rearranging some more and solving for a gives:
a= (mg-F)/(m + M)
The Tension on Block A then becomes M(mg-F)/(m + M) = 8.8 N
So then in order for Block A to resist motion, the Friction between it and the table must be equal to T, so F=T=8.8 .20 * (4.489 + C) * 9.8 = 8.8 but then the mass of C ends up being zero. Where did I go wrong?
There are two blocks, A (44 N) and B (22 N) connected via a rope that stretches over a pulley. Block A is resting on a table, and block B is hanging over the edge. Block C is positioned on top of Block A.
The question reads: "Given that the coefficient of static friction between Block A and the tabel is 0.20, what is the minimum weight of Block C to keep Block A from sliding off the table?"(The answer that is given is 66 N, but I can't seem to get this)I tried starting off with:
Block A=M=44N Block B=m=22N
(let F stand for the friction force)
T-F=Ma and mg-T=ma
Solving for Tension and setting them equal gives:
mg-ma=F+Ma
Rearranging some more and solving for a gives:
a= (mg-F)/(m + M)
The Tension on Block A then becomes M(mg-F)/(m + M) = 8.8 N
So then in order for Block A to resist motion, the Friction between it and the table must be equal to T, so F=T=8.8 .20 * (4.489 + C) * 9.8 = 8.8 but then the mass of C ends up being zero. Where did I go wrong?
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