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Showing posts with label A-LEVEL MATHEMATICS-MECHANICS. Show all posts
Showing posts with label A-LEVEL MATHEMATICS-MECHANICS. Show all posts
One bottle holder:how it works?
No, it's not magic. Its equilibrium like the man walking on a tight rope, but far less dangerous.
There is a physical quantity which ranks how effective is a force in rotating an object. It is called moment or torque of the force.Torque (τ) is defined as the magnitude of the force (F) times the distance (d) between the force's line of action and the pivot (the point, or shaft on which the object turns or oscillates)
If an object does not rotate then the total or net torque (τnet) of all the forces exerting on it should be zero.It's worth emphasising that the net torque should be zero for all possible pivots, since if it was not for one point then definitely there would be rotational motion around this very point).
Three forces are exerted on our one bottle holder system:
the wood's weight (w1), the weight of the bottle and its content (w2) and the ground reaction force N.
The point of application of the weight of an object is called center of gravity. We can find where the center of gravity of an object is. For example, you can determine where your pen's center of gravity using balance.Therefore we can assume that we know the centers of gravity for both the bottle and the wood and they are P and O respectively.Although the center of gravity is well defined and unique for an object, the point of application of the ground reaction force N can be defined but it is not unique. Let's call it Q and try to find where it is.
Since for equilibrium, the total torque should be zero for any pivot, let's refer to the point Q
- The moment of w1 is -w1d (negative since it contributes to clockwise rotation -with respect of Q).
- The moment of w2 is +w2d (positive since it contributes to anticlockwise rotation -with respect of Q).
- The moment of N is zero since its distance from Q is zero.
Therefore using equation 1 we can determine the position of Q.
What if the position of Q , as it is found above, does not belong to the area of touch between the wood and the table?
The answer:
Our one bottle holder will tip over!
Why?
Because N is a contact force and as such it should have a point of application within the area of touch. If the appropriate location for Q is outside the basis of our construction then we deduce that the current 'area of touch' is not enough to prevent our object from tipping over.
The object does not tip over!
The facts:
1. The main properties of a given force
are:
the point of application,
-the magnitude,
- the line of action and
-the sense.
Let us refer to these properties that are usually confused or vaguely understood. These are the point of application and the line of action.
the point of application,
-the magnitude,
- the line of action and
-the sense.
Let us refer to these properties that are usually confused or vaguely understood. These are the point of application and the line of action.
The point of
application is the exact location at which a force is applied to a body.
A force can be seen as a segment of an indefinitely long
line. To each force is associated a characteristic line, which is referred to
as its line of action.
2. A contact force is a force that acts at
the point of contact between two objects. In the figure the ground reaction
force which is analysed to its components R and F, is a contact force.
If there is no contact, there is no such a force. Since the magnitude of F (friction) is proportional to R, we say that when an object looses contact with the ground N=0 (and thus F=0)
Hint: when we are asked to find a condition so as an object does
not tip over, we start stating that N≥0 and then we apply the conditions for equilibrium.
By working these equations out, we make R the subject and then we demand N to
be non negative.
3. Conditions for equilibrium
·
An object
is in translational equilibrium when the sum of all the forces acting on the object equals zero.
In translational equilibrium, an object is either not moving, or moving at a constant velocity.
An object is in rotational equilibrium when the sum of all the external torques (moments) acting on it equals zero. In rotational equilibrium, an object either will not be moving, or moving at a constant angular velocity
In translational equilibrium, an object is either not moving, or moving at a constant velocity.
An object is in rotational equilibrium when the sum of all the external torques (moments) acting on it equals zero. In rotational equilibrium, an object either will not be moving, or moving at a constant angular velocity
4. Moment or torque of a force is a measure of that force's tendency
to cause a rotational acceleration in the same way that a force causes a linear
acceleration. In its simplest form, where a force of magnitude F is
acting a perpendicular distance d from the point of rotation, the
torque about that point is given by: τ=Fd
Example
A man is standing on a board which
has length L=4m and weight W1=150N. The board is supported at points A and B on
two trestles. The distance between these points is 2d=2m. The man weights w2=700N.
Calculate the maximum distance from the centre of the board at which the man
can stand without the board tips over.
In order to apply the condition for rotational equilibrium we
took the moments about B:
By applying the condition for translational equilibrium we
have
Since we don’t want the board to tip over:
Challenge
hints:
To start off apply the condition for rotational equilibrium taking
the moments around G
Then we resolve forces to their components perpendicular and parallel to the plane (advice the provided figure)
Then we try to solve the above formed equations with respect
of N1 and N2.
We claim that in order to avoid tipping over N1≥0 and N2≥0
Very useful will be the compound angles formulas.
Very useful will be the compound angles formulas.
We have to demonstrate good algebraic skills to reach to the desired
result.
Good luck.
A LEVEL MATHS MECHANICS 3 PLUS
The following exercise combines springs, oscillations, stability of a rigid body and rolling without slipping. It was the fourth exercise in the 2016 paper of physics in Greek National University entrance exams.
A block Σ having mass m=1 Kg is attached to the lower end of an ideal elastic spring of constant (stiffness) K=100 N/m. The upper end of the spring is attached to a firm point at the top a plane inclined at an angle φ=300 to the horizontal . The part BΓ of the inclined plane is smooth.
A block Σ having mass m=1 Kg is attached to the lower end of an ideal elastic spring of constant (stiffness) K=100 N/m. The upper end of the spring is attached to a firm point at the top a plane inclined at an angle φ=300 to the horizontal . The part BΓ of the inclined plane is smooth.
A solid homogeneous cylinder having
mass M=2Kg and radius R=0.1m is connected to the Block through a
massless and unstrechable cord. The cylinder's axis is horizontal.
The cord and the spring's axis are parallel to the inclined plane.
The system of all the bodies is initially at rest.
At t=0 the cord is broken. Block P
starts a simple harmonic oscillation while cylinder starts rolling
without slipping.
B1. write the formula that gives the
restoring force as a function of time. Take as positive the direction
towards the bottom of the inclined plane.
answer
answer
B3. Calculate the cylinder's kinetic
energy rate of change at t=3s.
The following information is given:
- the acceleration of gravity g=10m/s2
- the cylinders moment of inertia: I=0.5MR 2
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