phy唔識計...

1. Horizontal force component = F.sin(x)
hence, use force = mass x acceleration
F.sin(x) = 4800 x 4 ------------------ (1)
Verical force component = F.cos(x)
hence, for equilibrium in the vertical direction
F.cos(x) = 4800g ---------------- (2)
where g is the acceleration due to gravity, taken to be 10 m/s2

(1)/(2): tan(x) = 4/g = 0.4
i.e. x = 21.8 degrees
hence, F = 4800 x 4/sin(21.8) N = 51700 N

2. (a) Horizontal force = 75 x 7 N = 525 N
Verical force = 75g = 750 N, where g is the acceleration due to gravity
(b) Magnitude of resultant force
= square-root[525^2 + 750^2] N = 915 N
Let a be the angle which the force makes with the horizontal, hence,
tan(a) = 750/525
a = 55 degrees
The resultant force is at angle 55 degrees above the horizontal.

第1題o個道F = 4800 x 4/sin(21.8) N = 51700 N 點泥架??
係泥自 a=f/m ??

來自 F=ma,,,
F 是用 3角原理~要找的F為原F的sin(21.8deg)...所以要乘 sin21.8deg

(1) Let F be the thrust.
Let X be the angle the thrust makes with the vertical.
Now consider vertical direction. As the helicopter is in equlibrium (no vertical motion), the net force is zero. Hence,
FcosX = mg-----(1)
Now in the horizontal direction, the acceleration is 4 (m/s^2).
The net force = FsinX = ma (Newton's secondl law) ----(2)
(2)/(1) = tanX = a/g, X = arctan(a/g), a = 4 (m/s^2)
Also, (1)^2 + (2)^2 = F^2 = m^2 (a^2 + g^2)
F = (m^2 (a^2 + g^2))^(1/2)
Hence, X = 21.08 (degrees), taking g = 10(m/s^2)
F = 51700(N)

(2) (a) Consider vertically, the net force is zero newton as there is no acceleartion.
Hence, Fy (vertical component of force) = mg = 750 (N)
Horizontally, the acceleration = 7 (m/s^2). Hence, net force
= Fx = ma = (75)(7) = 525(N)

(b) Magnitude of the force = (Fx^2 + Fy^2) ^(1/2) = 915(N)
Let X be the angle of the force made with the horizontal,
tan X = Fy / Fx
X = 55 (degrees)