Answer:
fr = 269.3 N
Explanation:
Let's use Newton's second law, for this it is good to see the attached diagram,
X axis
fr -F2 = 0
fr = F2
Y Axis
N-W = 0
We must include the rotation balance, place the rotation point at the bottom of the ladder and take the positive counterclockwise turns.
Σ τ = 0
F2 x -W y / 2 = 0
We look for x and y with trigonometry
sin 70 = y / L
cos 70 = x / L
y = L sin70
x = L cos 70
We substitute and calculate F2
F2 L cos 70 = W L sin 70 / 2
F2 = mg/2 tan 70
F2 = 20 9.8/2 tan 70
F2 = 269.3 N
From the first equation (x axis)
fr = F2
fr = 269.3 N
Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
![\sum Fx = ma_x \ and \ \sum Fy = ma_y](https://tex.z-dn.net/?f=%5Csum%20Fx%20%3D%20ma_x%20%20%5C%20and%20%5C%20%5Csum%20Fy%20%3D%20ma_y)
The magnitude of the net force which is also known as the resultant will be expressed as ![R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}](https://tex.z-dn.net/?f=R%20%3D%5Csqrt%7B%28%5Csum%20Fx%29%5E2%20%2B%20%28%5Csum%20Fx%20%29%5E2%7D)
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
![a_x = \frac{d^2 x }{dt^2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%5E2%20x%20%7D%7Bdt%5E2%7D)
![a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdx%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%286t%5E%7B2%7D-4%20%20%29%5C%5C%5C%5Ca_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2812t%20%20%29%5C%5C%5C%5Ca_x%20%3D%2012m%2Fs%5E%7B2%7D)
Similarly,
![a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Cfrac%7Bdy%7D%7Bdt%7D%20%29%5C%5C%20%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%285t%5E%7B3%7D%20%2B6%20%29%5C%5C%5C%5Ca_y%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2815t%5E%7B2%7D%20%20%20%29%5C%5C%5C%5Ca_y%20%3D%2030t%5C%5Ca_y%20%5C%20at%20%5C%20t%3D%202.15s%3B%20a_y%20%3D%2030%282.15%29%5C%5Ca_y%20%3D%2064.5m%2Fs%5E2)
![\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N](https://tex.z-dn.net/?f=%5Csum%20F_x%20%3D%203.15%20%2A%2012%20%3D%2037.8N%5C%5C%5Csum%20F_y%20%3D%203.15%20%2A%2064.5%20%3D%20203.18N)
![R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B37.8%5E2%2B203.18%5E2%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B1428.84%2B41%2C282.11%7D%5C%5C%20%5C%5CR%20%3D%20%5Csqrt%7B42.710.95%7D%5C%5C%20%5C%5CR%20%3D%20206.67N)
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
Answer:
155.17N.
Explanation:
The magnitude of the net force is expressed as;
F = mv²/r where;
m is the mass
v is the velocity of the airplane
r is the radius of the loop
Given
m = 95kg
v = 70m/s
r = 3km = 3000m
Required
Magnitude of the net force
F = 95*70²/3000
F = 95*4900/3000
F = 95*49/30
F = 4655/30
F = 155.17N
Hence the magnitude of the net force on the 95 kg pilot at the bottom of this loop is 155.17N.
The main reason why it's very dangerous to open compress gas tank is it can cause major disaster not only in human but also in plants and animals. Compress gas is flammable. And it's like a bomb the damage of gas tank is very catastrophic. It is very harmful if mishandled.
As we know larger the area of contact lesser the pressure. So, in order to reduce the pressure heavy vehicles have broad tyres to increase the area of contact with the ground. Heavy vehicles have broad tyres because broad tyres have large area of contact and less pressure on the ground.
mark me brainliesttt pls :)))