The force, in newtons, exerted by each of the 10 braces is 2.135 x 10⁴ N.
<h3>What is force?</h3>The force is defined as the shear stress or pressure applied per unit area.
F = P/A
Given is a 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure(attached). The force is exerted by each of the 10 braces, if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths.
Considering the pivot at the base of wall.
From the equilibrium of forces, we have
r₁ x Fwind = r₁ x Fbsinθ
Put the values, we get
Fb = Fwind /10sin35°.............(1)
The wind force is also given by
Fwind = Horizontal force or pressure x Area
Fwind = F/A x hw
=655 x 17 x 11
F wind = 122,485 N
From equation (1), we have
Fb = Fwind /10sin35°
Fb = 122,485 /10sin35°
Fb = 21,354.6080 N
0r Fb = 2.135 x 10⁴ N
Thus, the force exerted on each of the 10 brace is 2.135 x 10⁴ N.
Learn more about force.
#SPJ1
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
= 454.55 g/cm3
I'm not too sure since the graduated cylinder was missing and I really don't know how to do it then. But give this a shot. Are you sure it wasn't a graduated cylinder, because I have no idea what that means
to find i1, i2, and i3 we need to find the total current.
to find the total current, you need to find the total resistance
you're already given the total voltage, Vs
to find Rtotal, start from the resistors furthest from the voltage source.
R3 and R4 are in series so
Rtotal= R3+R4 = 6+3 = 9 ohms
9 ohms is now in parallel with R2 so,
Rtotal= + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms
6 ohms is in series with R1 so
Rtotal= 4+6=10 ohms
itotal= (\frac{Vtotal}{Rtotal})
= 120 v/10 ohms = 12 A
i total = i1 because all the current flows through it
i1= 12A
so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.
we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms
since r3 and r4 are in series, the same current will flow through them
r3+r4 = 9 ohms
r2= 18 ohms
so the current in r2 will be half that of R3 & R4 (V=IR)
using the current divider rule
i2= 12A x (6 ohms/18 ohms)= 4 A
i3= 12A x (6 ohms/9 ohms) = 8 ohms