Total Belt Length = 2*L1 + (length around pulley 1) + (length around pulley 2)

angle a = arcsin(0.75/4) = 10. 8 degrees

L1 = 4*cos(10.8) = 3.93 M

angle b = 90 - 10.8 = 79.2 degrees = 1.38 radians

Length aroud pulley 1 = 2*(PI - 1.38) * (1.0M) = 3.52 M

Length aroud pulley 2 = 2*(PI -(a+90 degrees ) * r2

= 2*(1.38 rad)*0.25 = 0.69 M

Total Belt Length = 2*(3.93) + 3.52 + 0.69 = 12.07 M

1.29)

1. Find the dimensions of F in F = ma

    F = [M][[L/(T²)]] = [MLT^-2]

2. In R = cv +ag, R is force, v is velocity, and g is acceleration. Find the dimensions of c and a

c = MT^-1  

a = M

Then:

[MLT^-2] = [MLT^-2]+[MLT^-2]

3. Determine the dimensions of P and I in the following equation given d is a length, m is a mass, v is a linear velocity, and w is an angular velocity (Note: angular velocity ir radians per second ,which has dimensions 1/T -- a radian is not a length. If you think it is, tell me what unit of length measures 1 radian).

PD = 0.5mv²+0.5Iw²  

P[L] = [M][L²T^-2]+I[T^-2]

P = [MLT^-2]

I = [ML²]

4. Given that A is length and t is time, determine the dimensions of y,b,a and c in the following equation:

  y = Ae^-btcos([1/(-a²bt)]+c)

The argument of the exponent and trig function must be dimensionless, so we'll deal with them separately. Since those functions will also return dimensionless quantities, y and A must have the same dimension, [L]. Next, look at the exponential term. If t is [T], and -bt is dimensionaless, then b must be [1/T]. Finally, let's look at the cosine argument. If the entire argument is dimensionless, then each term in the equation must be dimensionless via the dimensionally homogeneous principle. Therefore, c is dimensionless. In the other term, the bt is the denominator is dimensionless, and since no other factor has a dimension, a must be dimensionless as well.

5. In the equation below, m is mass, vi is velocity, R is force, d is length, and a is acceleration. Is the equation dimensionally correct?

  mv² = rd+ma*cos(Q)

[M][LT^-1]² = [MLT^-2][L]+[M][LT^-2]

[ML²T^-2] = [ML²T^-2]+[ML²T^-2]+[MLT^-2]

NO, the equation is not dimensionally correct.