if the mass of the objects stays the same and their distance from eachother decreases, then the force of gravity between the two
1 answer:
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a) Angular acceleration:
b) Weight: conterclockwise torque, reaction force: zero torque
Explanation:
a)
In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.
The only force producing a torque on the pencil is the weight of the pencil, of magnitude
where m is the mass of the pencil and g the acceleration of gravity.
However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:
where is the angle of the rod with respect to the vertical.
The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is
where L is the length of the pencil.
The relationship between torque and angular acceleration is
(1)
where
is the moment of inertia of the pencil with respect to its end.
Substituting into (1) and solving for , we find:
And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when is
b)
There are only two forces acting on the pencil here:
- The weight of the pencil, of magnitude
- The normal reaction of the hand on the pencil, R
The torque exerted by each force is given by
where F is the magnitude of the force and d the distance between the force and the pivot point.
For the weight, we saw in part a) that the torque is
For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.
Therefore:
- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)
- Reaction force: zero torque
Answer:
A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N
B. Fn= 6.78 *10⁻⁵ N
C. α= 32.4° counterclockwise with the positive x+ axis
Explanation:
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Equivalences
1nC= 10⁻⁹C
1cm = 10⁻²m
Known data
k= 9*10⁹N*m²/C²
q₁= -2.65 nC =-2.65*10⁻⁹C
q₂= +2.00 nC = 2*10⁻⁹C
q₃= +5.00 nC= =+5*10⁻⁹C
* 10⁻²m = 4.9678* 10⁻²m
(d₁₃)² = 24.68*10⁻⁴m²
d₂₃ = 3.2 cm = 3.2*10⁻²m
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.
The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.
Magnitudes of F₁₃ and F₂₃
F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)
F₁₃ = 4.8 *10⁻⁵ N
F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)
F₂₃ = 8.8 *10⁻⁵ N
x-y components of F₁₃ and F₂₃
F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N
F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N
F₂₃x = F₂₃ = +8.8 *10⁻⁵ N
F₂₃y = 0
x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)
Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N
Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N
Fn magnitude
*10⁻⁵ N
Fn= 6.78 *10⁻⁵ N
Fn direction (α)
α= -32.4°
α= 32.4° counterclockwise with the positive x+ axis
