HOMEWORK CHAPTER 5

 


1) A book on a table has a weight of 9.0 N. A force Ps = 3.0 N is required to start it sliding. But to keep it sliding at a constant speed, a minimum force Pk = 2.0 N is required. Find:

a) the coefficient of static friction and b) the coefficient of kinetic friction.

Ans. a. 0.33 b. 0.22 

21) An object is moving on a curved path at a linear speed v = 30.0 m/s in a circle of radius r = 2.00 m. The object’s mass is m = 0.100 kg. Find the centripetal force.

Ans. 45.0 N 

29) The mass of the earth is, ME = 5.98 x 1024 kg, and the mass of the moon, MM = 7.36 x 1022 kg. The distance between their centers is 3.82 x 108 m. How far from the center of the earth is the center of mass of the Earth-Moon system?

Ans. 4.64 x 106 m. [the radius is the earth is 6.37 x 106 m, placing the center of mass inside the earth.]


SOLUTIONS 

(1)

(a)

The coefficient of static motion is described as the maximum ratio obtained between applied force and the normal force of a body in a static position. In other words, is the maximum resistive force provided by a body counteract the applied force. So that body can continue in its static position. Mathematically, the Coefficient of Static Friction is the ratio between applied force and normal force. It is represented as follows:

mus= *Fs/*FN, where *Fs and *FN are the applied force and normal force respectively.

Normal force (*FN) = mgN

In this case, the normal force (N), is its weight.

N=mg= 9N, where m is the mass of the object and g is the acceleration of gravity. Furthermore, Ps is the applied force. Therefore, its Coefficient of Static Motion becomes:

mus= *Fs/*FN (3.0N/9.0N)= 0.33

(b)

The Coefficient of Kinetic Friction is described as the ratio between the friction force, and the normal force, and it is given by:

mu=F/N, where F is the friction force and N is the normal force.

mu=2.0N/9.0N = 0.22

(21)

Centripetal force (Fc) is given by Fc=mac, and ac is the centripetal acceleration.

ac=v2/r=r (angular frequency)2

Therefore, Fc=0.100Kg*1/2.0m* (30.0m/s) 2= 45.0N

(29)

For simplification of calculations, let the origin be the center of the Earth. For that reason, subscripts E and M, denote Earth and Moon respectively. y is the distance between the center of the Earth-Moon System and Earth’s center.

Xcm= (mMXM + mEXE)/ (mM + mE)

Since Earth’s center is the origin mEXE becomes 0.

Distance between Earth’s center and Earth-Moon System (y) = (mMXM + 0)/ (mM + mE)

y=mMXM / (mM + mE)

3.82*108m 7.36*1022/ (7.36*1022 + 5.98*1024) m

y=4.64*106m 

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