class 051024

The force experiment is at http://www.vhcc.edu/ph1fall9/experiments/rubber_band_calibrations_and_vector_expts/force_vectors.htm.  Calibrate rubber bands by hanging dominoes from the chain.  Force units will be dominoes.

What are the x and y components of a vector of length 7 meters which make an angle of 20 degrees with the positive x axis?

Let L stand for this vector.  Then

Lx = L cos(theta) = 7 m cos(20 deg) = 6.6 m

Ly = L sin(theta) = 7 m (sin(20 deg)) = 2.4 m

What are the four possible angles, as measured counterclockwise from the positive x axis, of a vector of magnitude 7 which makes an angle of 20 degrees with the y axis?  What are the possible x and y components of this vector?

The vector could be in the first quadrant, 20 deg clockwise from the positive y axis.  The angle of this vector measured counterclockwise from the positive x axis is 90 deg - 20 deg = 70 deg.

In this case we would have

vx = 7 cos(70 deg) = 2.4

vy = 7 sin(70 deg) = 6.6.

The vector could be in the second quadrant, 20 deg counterclockwise from the positive y axis.  The angle of this vector measured counterclockwise from the positive x axis is 90 deg + 20 deg = 110 deg.

In this case we would have

vx = 7 cos(110 deg) = -2.4

vy = 7 sin(110 deg) = 6.6.

 

The vector could be in the third quadrant, 20 deg clockwise from the negative y axis.  The angle of this vector measured counterclockwise from the positive x axis is 270 deg - 20 deg = 250 deg.

In this case we would have

vx = 7 cos(250 deg) = -2.4

vy = 7 sin(250 deg) = -6.6.

 

The vector could be in the fourth quadrant, 20 deg counterclockwise from the negative y axis.  The angle of this vector measured counterclockwise from the positive x axis is 270 deg + 20 deg = 290 deg.

In this case we would have

vx = 7 cos(290 deg) = 2.4

vy = 7 sin(290 deg) = -6.6.

A 5 kg mass is released from the top of an incline 7 meters long, which is inclined at an angle of 20 degrees with horizontal.

• What are the horizontal and vertical displacements of the mass, as measured from its starting position, as it rolls to the bottom of the incline?
• If we have the origin at the top of the incline, x and y axes horiz to right and vertical upward, incline sloping off down into the 3d quadrant then the vector displacement from top to bottom is has magnitude 7 m and angle 200 deg.  This vector has x and y components

Lx = L cos(theta) = 7 m cos(200 deg) = -6.6 m and

Ly = L sin(theta) = 7 m sin(200 deg) = -2.4 m.

Horizontal and vertical displacements are -6.6 m and -2.4 m, respectively.

• What is the gravitational potential energy change of the mass?
• Quick and dirty calculation:  The weight is 5 kg * 9.8 N = 49 N, displacement is 2.4 m downward so PE decreases by 49 N * 2.4 m = 117 Joules.
• More precise and rigorous setup: 

PE change is work done by the mass against the conservative force, in this case gravity.

The force exerted by gravity ON the mass is downward, in the negative y direction.  The magnitude of this force is weight = m g = 5 kg * 9.8 m/s^2 = 49 N.

So the force exerted BY the system against gravity is equal and opposite, upward in the positive y direction.  So the force exerted against gravity is +49 N.

The displacement of the mass is in the negative y direction, so `dy = -2.4 m.

The change in PE is therefore F * `dy = 49 N * (-2.4 m) = -117 J.

• What is the work done by gravity on the mass as it rolls down the incline?
• Gravity exerts its 49 N force in the downward direction.  The vertical displacment of the mass is downward.  So the work done by gravity ON the mass is

F * `dy = -49 N * (-2.4 m) = +117 J.

The force and the displacement are in the same direction, so the work is positive.

Gravity does positive work ON the mass.

• What is the work done by the mass against gravity as it rolls down the incline?
• Gravity does positive work ON the mass, so the mass does negative work against gravity.  As also seen before, this results in work -117 J being done BY the mass against gravity.
• In what direction is this system in equilibrium?
• All the acceleration is along the incline.  There is no motion, and hence no acceleration, perpendicular to the incline.  The system is in equlibrium in a direction perpendicular to the incline.

If there is a constant frictional force of 2 Newtons opposing the motion of the mass down the incline, and if the mass starts from rest, then

• How much work is done against friction as the mass rolls down the incline?

The work is `dW_frict = f_frict * `ds, where `ds is the displacement along the line of the friction.

In the direction of the friction the displacement is 7 m down the incline.

Choose down the incline as the positive direction.

The direction of the frictional force ON the system is up the incline therefore negative.

The direction of the force exerted BY the system against friction is equal and opposite, therefore positive.

The work done BY the system against friction is therefore

`dW_frict_BY = +2 N * (+7 m) = +14 Joules.

• How much work is done by friction as the mass rolls down the incline?
• Still using down the incline as the positive direction, the force exerted ON the system by friction is -2 N and the work done by the frictional force is

`dW_frict_ON = -2 N * (+7 m) = -14 Joules.

• What is its KE change as it rolls down the incline?
• Intuitively, there is a 117 J loss of PE, which will all go into KE if nothing interferes.  However friction interferes, to the tune of a 14 Joule loss, so the KE increases by only 117 J - 14 J = 103 J.
• What is its KE at the bottom of the incline?
• This will be 103 J, since it starts from rest and the change in 103 J.
• What is its speed at the bottom of the incline?
• We have 1/2 m v^2 = KE so that v = +-sqrt( 2 * KE / m) = +_ sqrt( 2 * 103 J / (5 kg) ) = +- sqrt( 41.2 m^2 / s^2) = 6.5 m/s.
• What are the x and y components of its velocity at the bottom of the incline?
• The velocity at the bottom of the incline is 6.5 m/s, along the direction of the incline.  The direction of the incline is at 200 deg to the positive x axis.

So the x and y components of the velocity are

vx = 6.5 m/s * cos(200 deg) = -6.1 m/s and

vy = 6.5 m/s * sin(200 deg) = -2.2 m/s.

How does the above situation of the mass on the incline fit into each of the following formulations of the work-energy theorem?

• `dW_noncons_ON = `dPE + `dKE
• `dW_noncons_ON is the -14 J of work done by friction ON the system.

`dPE is the -117 J of PE change.

`dKE is the 103 J of KE increase.  The equation becomes

-14 J = -117 J + 103 J,

which is true.

• `dW_noncons_BY + `dPE + `dKE = 0
• `dW_noncons_BY = +14 J done BY the system against friction.
• `dPE = -117 J.
• `dKE = 103 J.
• The equation reads
• +14 J - 117 J + 103 J = 0,
• which is true.
• `dW_net_ON = `dKE