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
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
where are the magnitudes of the tensions in ropes 1 and 2, respectively;
where and
.
Eliminating , we have
Solve for .
Answer:
50 N.
Explanation:
On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:
The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:
In conclusion, the normal force acting on the block is 50 N.
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s
2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration
3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.
So the maximum acceleration on the block is
4)As the box slides, it is now subjected to kinetic friction, which is
So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2
Answer:
If the frequency of the source is increased the current in the circuit will decrease.
Explanation:
The current through the circuit is given as;
Where;
V is the voltage in the AC circuit
Z is the impedance
Where;
R is the resistance
is the inductive reactance
= ωL = 2πfL
where;
L is the inductance
f is the frequency of the source
Finally, the current in the circuit is given as;
From the equation above, an increase in frequency (f) will cause a decrease in current (I).
Therefore, If the frequency of the source is increased the current in the circuit will decrease.