nineeyes
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I attached the problem and diagram.
This problem has been bothering me a bit. A friend of mine told me I could solve for the stopping distance by just looking at the crate.
I got
x: [tex]\mu*N_w = M*a[/tex]
y: [tex]N_w-W_w=0[/tex]
and from this got
[tex]a=-9.66\frac{ft}{s^2}[/tex]
I used the acceleration and given inital velocity of 10 ft/s and final velocity of 0 ft/s and solved for the stopping distance through
[tex]s=\frac{-v_0^2}{2*a}[/tex]
Assuming I approached this correctly, I wasn't sure how to check if it doesn't tip. I thought that I could just take the moment about a point and solve for [tex]N_B[/tex], I did
[tex]\sum M_A=N_B*7-W_G*4+W_W*3-a*3=0[/tex]
and got
[tex]N_B=210.15[/tex] lbs. [tex]\therefore[/tex] it does not tip
Is that right? I also was wondering if I have to consider frictional force in the moment equation.
Thanks in advance.
This problem has been bothering me a bit. A friend of mine told me I could solve for the stopping distance by just looking at the crate.
I got
x: [tex]\mu*N_w = M*a[/tex]
y: [tex]N_w-W_w=0[/tex]
and from this got
[tex]a=-9.66\frac{ft}{s^2}[/tex]
I used the acceleration and given inital velocity of 10 ft/s and final velocity of 0 ft/s and solved for the stopping distance through
[tex]s=\frac{-v_0^2}{2*a}[/tex]
Assuming I approached this correctly, I wasn't sure how to check if it doesn't tip. I thought that I could just take the moment about a point and solve for [tex]N_B[/tex], I did
[tex]\sum M_A=N_B*7-W_G*4+W_W*3-a*3=0[/tex]
and got
[tex]N_B=210.15[/tex] lbs. [tex]\therefore[/tex] it does not tip
Is that right? I also was wondering if I have to consider frictional force in the moment equation.
Thanks in advance.