Study Guide 10 - Mechanics of Biological Systems: Forces and Motion

\(\mathrm{F = ma \\ 50 = 20a \\ a = 50/20 = 2.5 \;m/s^2}\)

Is there any unbalanced force in the vertical direction?

No. 

Is the force that accelerates the block: \(\mathrm {F}\)  or  \(\mathrm {F\; \cos\; 36.9}\)  or  \(\mathrm {F\; \sin \;36.9 ?}\)

The horizontal component of \(\mathrm {F = 50\; \cos \;36.9 = 50 \times 0.8 = 40\; N}\)
\(\mathrm {F = ma}\)
\(\mathrm {a = F/m = 40/20 = 2.0 m/s}\)

First find the required acceleration.

\(\mathrm {v = v_0 + at \\ a = (0 - 20)/5 = -4.0 m/s^2}\)

\(\mathrm {F = ma = 10 \times (-4.0) = -40\; N}\)

\(\mathrm {On \; Earth, W = mg = 50.0 \times 9.80 = 4.90 \times 10^2 \; N}\)

On Mars the effective \(\mathrm {g}\) is less.

Since gravitational attraction varies with the product of the masses invomved and inversely as the square of the distance (in this case the radii of the planets) then

\(\mathrm {g_{Mars} = g_{Earth}(M_{Mars}/M_{Earth})(R_{Earth}/R_{Mars})^2 \\ = (9.8) \times (6.46\times10^{23}/6.0\times10^{24})(6.37\times 10^6/3.40\times 10^6)2 = 3.7\; m/s^2 \\ W = mg_{Mars} = 50 \times 3.7 = 185\; N}\)

The point where F attaches to the cable is a point which is in equilibrium, i.e., \(\mathrm {\sum F = 0.}\)

How many forces are acting at that point?

There are 3 forces acting at the point. These are: F and the tension T in the cable acting toward the car and toward the tree.

\(\mathrm {\sum F_x = 0 \\ T_2\; \cos \; \theta - T_1 \; \cos \; \theta = 0 \\ Therefore,\; T_1 = T_2 = T}\)

\(\mathrm {\sum F_y = 0 \\ F - 2T \; \sin \; \theta = 0}\)

Remember that the angle is small so:
\(\mathrm {\sin \; \theta = \theta = 0.1/25 \\ T = F/2 \sin\theta = 500/(2 \times 0.1/25) = 6.25\times 10^4 \;N}\)

Vertical components (y)

\(\mathrm {30\; \sin \;60 - R\; \sin \theta = 0}\)
\(\mathrm {R\; \sin \theta = 30\; \sin\; 60 = 26  \qquad   [1]}\)

Horizontal components (x)

\(\mathrm {30\; \cos\; 60 + R\; \cos \theta = 20}\)
\(\mathrm {R\; \cos \theta = 20 - 30\; \cos\; 60 = 5   \qquad  [2]}\)

Divide [1] by [2]
\(\mathrm{\tan \theta = 26/5 = 5.2}\)
\(\mathrm {\theta = 79.1^\circ}\)

From [1]
\(\mathrm {R = (30\; \sin \;60)/(\sin\; 79.1) = 26.5 \;N}\)

We must first make a free-body diagram of the packing case.
The forces acting on the case are:

• The 50 N force at 30º which can be divided into 2 components:
  a: 50 cos 30 to the right
  b: 50 sin 30 up
• The force of friction acting to the left.
• The normal reaction of the floor acting up.
• The weight of the case 200g acting down.

The free-body diagram is therefore:

What is the acceleration of the packing case in the vertical direction?
(a) \(\mathrm {0.100\; m/s^2}\) - No. That is the horizontal acceleration.
(b) \(\mathrm {9.8 \;m/s^2}\) - No. That is the acceleration of an object in free-fall.
(c) 0 - Correct! The case does not accelerate in the vertical direction.

Therefore in this direction the sum of the forces is zero.
\(\mathrm {\sum F = 0}\)

\(\mathrm {N + 50\; \sin \; 30 - 200\; g = 0}\)
\(\mathrm{N = 1960 - 25 = 1935\; N }\)  - this is the answer to part b

The acceleration in the horizontal direction is \(\mathrm {0.100 \;m/s^2.}\) The sum of the forces in the horizontal direction is given by:

\(\mathrm {\sum F = ma}\)

\(\mathrm {50\; \cos \;30 - f = 200(0.100)\\ f = 43.3 - 20 = 23.3\; N}\)
 

\(\mathrm{N = 1960 - 25 = 1935\; N }\)

When an object moves at constant velocity there is no net force (Newton's 1st law).
Therefore the persons apparent weight, which is the force the elevator floor exerts upward on the person
equals the persons own weight acting downward, i.e.,
\(\mathrm {App\; Wt = mg = 60 \times 9.8 = 588 \;N}\)

If the apparent weight is F (the force the elevator floor exerts upward on the person) then
\(\mathrm {mg - F = ma}\) where \(\mathrm {a}\) is the persons acceleration

\(\mathrm {F = m(g - a) = 60 (9.8 - 1.6) = 492 \;N}\)

The component of force down the plane is \(\mathrm {mg \; \sin \; 37}\) and from Newton's second law this must equal \(\mathrm {ma}\)
\(\mathrm {mg\; \sin \; 37 = ma \\ a = g\; \sin \;37 = 9,\;8 \times 0.6 = 5.9 m/s^2}\)

There is no acceleration perpendicular to the plane so
\(\mathrm {N - mg\; \cos \;37 = 0}\)
\(\mathrm {N = 100 \times 9.8 \times 0.8 = 784\; N}\)

\(\mathrm {\sum F_y = 0 \\ N + 10\; \sin \;37 - mg = 0 \\ N = mg - 10\; \sin \;37 = 5(9.8) - 10(0.6) = 43\; N \\ f = \mu N = 0.1 \times 43 = 4.3 \;N}\)

\(\mathrm {\sum F_x = ma \\ 10\; \cos \;37 - f = ma}\)

\(\mathrm {a = (1/5)[10(0.8) - 4.3] = 0.74 \;m/s^2}\)

\(\mathrm {v = v_0 + at = 0 + 0.74 \times 5 = 3.7 \;m/s}\)

Perpendicular to the plane
\(\mathrm {\sum F = 0 \\ N - mg\; \cos \;37.1 = 0 \\ N = mg\; \cos \; 37.1 \\ f = \mu N = 0.2 \times 20 \times 9.8 \times 0.8 = 31.36 \;N}\)

Along the plane
\(\mathrm {\sum F = ma \\ 200 - mg\; \sin \;37.1 - f = ma \\ 200 - (20)(9.8)(0.6) - 31.36 = 20a \\ a = 51.04/20 = 2.55 \;m/s^2}\)

\(\mathrm{v^2 = v{_0}^2 + 2ax \\ 0 = 9^2 + 2a4 \\ a = -10.125 m/^2}\)

\(\mathrm {f = ma = \mu mg}\)

\(\mathrm {a = \mu g}\)

\(\mathrm {\mu = a/9 = 10.125/9.8 = 1.03}\)

Momentum is conserved in all collisions

Both the magnitude and the direction of the momentum are conserved.
Now conservation of momentum can be applied to the horizontal (x) and vertical (y) components of the momentum independently

 

Before Collision	After Collision
\(\mathrm{p_x = 2000 \times 30 \\ p_y = 1000 \times 15}\)	\(\mathrm {p_x = 3000\; v \cos\; \theta \\ p_y = 3000\; v\; \sin \theta}\)

In the x direction: \(\mathrm {3000\; v \sin \theta = 1000 \times 15}\)
In the y direction: \(\mathrm {3000\; v \cos \theta = 2000 \times 30}\)

Divide the first equation by the second:
\(\mathrm {\tan \theta = 15/2\times 30 = 0.25 \\ \theta = 14^\circ}\)

\(\mathrm {3000\; v \cos\; 14 = 2000 \times 30 \\ v = 20.6\; m/s}\)

The problem asks only for the work done by the applied force.
\(\mathrm {W = Fd = 100 \times 3 = 300 \;J}\)

In the frictionless fall of the 10 kg mass what is conserved, Energy or Momentum?

Energy

Potential energy is converted into kinetic energy.
\(\mathrm {mgh = ½mv^2 \\ v = (2gh)^½ = [2(9.8)5]^½ = 9.9 \;m/s}\)

In the subsequent collision with the 5 kg mass what is conserved, Energy or Momentum?

Momentum

Momentum is always conserved in a collision but energy may, or may not be conserved.

\(\mathrm {Momentum\; before\; = mv = 10(9.9) \\ Momentum\; after = MV = (10 + 5)V = 15V \\ 15V = 99 \\ V = 99/15 = 6.6 \;m/s}\)

\(\mathrm {Change\; in \;potential \;energy = mgh = 100(9.8)(40) = 39200 \;J}\)

\(\mathrm {Power = E/t = 3920/30 = 1.31\times 10^2\; J}\)​​​​​​​