- #1
Xiongster
- 8
- 0
Homework Statement
Hi, I have a problem here and I need some help on it. The problem is this:
block m[tex]_{1}[/tex] weighs 875 N. The coefficient of static friction between the block and the table is 0.24 and the angle theta is 25.0[tex]\circ[/tex]. Find the maximum weight of block m[tex]_{2}[/tex] for which block m[tex]_{1}[/tex] will remain at rest.
2. The attempt at a solution
So far I've drawn an FBD for m[tex]_{1}[/tex], m[tex]_{2}[/tex], and the knot that connects them.
For the block m[tex]_{1}[/tex], I have T[tex]_{1}[/tex] = [tex]M[/tex][tex]_{s}[/tex]N in the x-direction, and N = m[tex]_{1}[/tex]g for the y-direction.
For block m[tex]_{2}[/tex], I have T[tex]_{2}[/tex] = m[tex]_{2}[/tex]g in the y-direction.
For the knot, I have T[tex]_{1}[/tex] = T[tex]_{3}[/tex]*cos(25) in the x-direction and T[tex]_{2}[/tex] = T[tex]_{3}[/tex]*sin(25) in the y-direction.
Since T[tex]_{1}[/tex] = T[tex]_{3}[/tex]*cos(25), and T[tex]_{1}[/tex] = [tex]M[/tex][tex]_{s}[/tex]N, I set T[tex]_{3}[/tex]*cos(25) = [tex]M[/tex][tex]_{s}[/tex]N and solving for T[tex]_{3}[/tex], I get T[tex]_{3}[/tex] = ([tex]M[/tex][tex]_{s}[/tex]N)/(cos(25))
In the knot, we have T[tex]_{2}[/tex] = T[tex]_{3}[/tex]*sin(25). I plug in the T[tex]_{3}[/tex] derived earlier and T[tex]_{2}[/tex] = m[tex]_{2}[/tex]g into this equation to get m[tex]_{2}[/tex]g = [tex]M[/tex][tex]_{s}[/tex]*m[tex]_{1}[/tex]g*tan(25)
When I plug in numbers, the answer I get is about 98, and that just doesn't seem right to me. Did I do something wrong? Thanks for the help in advance.