Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and i
nclude appropriate units.
1 answer:
Complete Question
A force of F =460 lb. acts on the curved member as shown. Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and include appropriate units.(Figures Attached)
Answer:
Resultant Normal Force = 55. 02 lb.
Explanation:
In Order to find the resultant normal force we must first resolve the figure to find out the x and y components of the force across the cross-section at point B. Hence after resolving the figure we can find the resultant normal force by finding the vertical component of the force: ∑ F_y
The normal force acting at point B will be:
∑ F_y :-
N_F + 460(4/5)sin〖30〗^° - 460(3/5) sin〖60〗^° = 0
N_F + 184 - 239.02 = 0
N_F = 55. 02 lb.
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Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M