1. When and why should we use the Pattern option?
1 answer:
You might be interested in
Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
Answer:
Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:
Where is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.
The horizontal and vertical distances are, respectively:
The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find
Using this time in the horizontal distance, we find the Range or maximum horizontal distance:
Let's solve for
This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:
Or equivalently:
Given Vo=37 m/s and R=70 m
And