I have a problem that I need assistance with:
There is a well insulated rigid tank with 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, half of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 120-V source...
This is my work for solving part a) when a=2
Sum of the Moment about A = 20(2)-16(1.5)+[5.5(sin(30))+2(cos(30)]E
E= -8/[5.5(sin(30))+2(cos(30)] or 3.57 lbs at a 60 degree angle, which is what the book says.
Now how come when I plug in 7.5 in a, I can't...
Just figured I'd throw this up here. I was able to solve for the reaction at E when a=2 but when I tried for a=7.5, I wound up with something completely different. I was wondering what I'm not incorporating into the steps to solving this.
Also, it says determine the reactions at A and E. The...
In response to my above comment, I really realize that I overlook things too much. 3.94 I have the j and k of the force couple system (use the force that I obtained when breaking down the 220 N in the cross product. Duh.) Somehow I cant seem to get i. Any help would be appreciated.
If I could actually ask for help in another question, any help would be appreciated.
Excuse my chicken scratch work. I solved for the angle at which the force is acting at, which is 30 degrees in the xz plane. I solve for the actually distance from the x axis that said force was, which was...
I need help on part c of 3.71. I have the angle from b, which I believe I need to use. I originally thought that the way to set it up would be
86.2 = x(22.36)*sin(53.1)
^22.36 coming from the sqrt of 17.6^2+ 13.8^2
Any help on what I'm doing wrong would be appreciated.
I have a couple problems that I can't seem to get.
The first I'm really close, I can tell, because my professor did one that was similar in class. The answer the book gives is 250 lbs, I'm coming up with 225.
The second I think is similar to another problem he did in class but I'm not...