Form 4 Static 兩條

3a) Let T be the tension of the string, then:

Horizontal forces: T sin θ = R where R is the normal reaction of the wall

Vertical forces: T cos θ + f = W where f is the friction of the wall on the wall

But f = μR, we have:

T cos θ + μT sin θ = W ... (1)

Now, considering the moments about the contact point between the ball and the wall:

By the string = Th sin θ (Perp. distance from the contact point to the line of force of T is h sin θ)

By the ball = Wa

Therefore:

Th sin θ = Wa

T = Wa/(h sin θ)

Sub into (1):

T cos θ + μT sin θ = W

Wa/(h tan θ) + μWa/h = W

Wa + μWa tan θ = Wh tan θ

a + μa tan θ = h tan θ

tan θ = a/(h - μa)

b) With tan θ = a/(h - μa) = 1/μ, we have:

sin θ = 1/√(1 + μ2)

T = Wa/(h sin θ)

= Wa√(1 + μ2)/h

= W√(1 + μ2)/(2μ)

5a) As follows:

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Feb11/Crazystatic1.jpg

b) Let θ be the angle between the string and the rod, with

tan θ = 3/4, thus sin θ = 3/5 and cos θ = 4/5

Considering the moments:

Tl sin θ = (Mg/2)(3l/4) + (Mg)(l/2)

3T/5 = 7Mg/8

T = 35Mg/24

So the horizontal component of R is T cos θ 7Mg/6

Also the vertical component of R is given by:

3Mg/2 - T sin θ = 3Mg/2 - 7Mg/8 = 5Mg/8

So the magnitude of R is:

Mg√[(5/8)2 + (7/6)2] = (Mg√1009)/24

And the angle of inclination is given by:

tan-1 [(5/8)/(7/6)] = tan-1 (15/28)

1.(a) Let T be the tension in the string, R be the normal reaction at the point of contact between the shpere and wall, Ff be the frictional force between sphere and wall.

Consider vertical forces:
T.cos(theta) + uR = W
[u is the coefficient of friction and W is the weight of the sphere]
i.e. T.cos(theta) = W - uR --------------------- (1)

Consider horizontal forces,
T.sin(theta) = R ----------------- (2)
(2)/(1): tan(theta) = R/(W - uR) -------------- (3)

Take moment about the point where the string is fixed to the wall,
W.a = R.h
i.e. R = Wa/h
substititue the value of R into (3) and simplfy, you would get the answer.

(b) (1)^2 + (2)^2: T^2[cos^2(theta) + sin^2(theta)] = (W - uR)^2 + R^2
i.e. T^2 = (W - uR)^2 + R^2 ----------------- (4)
Using R = Wa/h = W/2u substitute into (4) and simplify, you would get the answer.

2.Let T be the tension in the string, R be the reaction at A which makes an angle of, say a, with the hroizontal.

Since the string makes an angle (theta) given by tan(theta) = (3/4l)/l = 3/4, i.e. (theta) = 36.87 degrees.
The equations involved are:
Vertical forces:
T.sin(36.87) + R.sin(a) = Mg/2 + Mg
where g is the acceleration due to gravity
i.e. T.sin(36.87) + R.sin(a) = 3Mg/2 ------------------ (1)

Horizontal forces:
T.cos(36.87) = R.cos(a) ---------------- (2)

Taking moment about point A,
[T.sin(36.87)].l = (Mg/2).(3l/4) + (Mg).(l/2)
i.e. T.sin(36.87) = 7Mg/8 ------------------ (3)
i.e. T = 1.458Mg

Using the values of T just found, together with equations (1) and (2), you should be able to find R and the angle a.