Two aluminum soda cans are charged and repel each other, hanging motionless at an angle. Which of the forces on the left can has
1 answer:
Answer:
the tension is the greatest force the system
Explanation:
For this exercise we use Newton's second law, in the equilibrium condition, this means that the acceleration is zero a = 0
X-axis (horizontal)
FE - Tₓ = 0
Fe = Tₓ
Y axis (vertical)
- W = 0
T_{y} = W
let's use trigonometry for the stress components
sin θ = Tₓ / T
Tₓ = T sin θ
cos θ = T_{y} / T
T_{y} = T cos θ
we substitute
FE = T sin θ
W = T cos θ
Since the sine and cosine function have values between 0 and 1, the voltage must be greater than the electrical force and greater than the weight of the body.
Therefore the tension is the greatest force the system
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Answer:
a) u_s=0.375
b )v=14.4 m/s
Explanation:
1 Concepts and Principles
Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:
∑F=0 (1)
2- Particle in Uniform Circular Motion:
If a particle moves in a circle or a circular arc of radius Rat constant speed v, the particle is said to be in uniform circular motion. It then experiences a net centripetal force F and a centripetal acceleration a_c. The magnitude of this force is:
F=ma_c
=m*v^2/R (2)
where m is the mass of the particle F and a_c are directed toward the center of curvature of the particle's path.
<em>3- The magnitude of the static frictional force between a static object and a surface is given by:</em>
f_s=u_s*n (3)
where u_s is the coefficient of kinetic friction between the object and the surface and n is the magnitude of the normal force.
Given Data
R (radius of the car's path) = 170 m
v (speed of the car) = 25 m/s
Required Data
- In part (a), we are asked to find the coefficient of static friction u_s that will prevent the car from sliding.
- In part (b), we are asked to find the speed of the car v if the coefficient of static friction is one-third u_s found in part (a).
Solution:
see the attachment pic
Since the car is not accelerating vertically, we can model it as a particle in equilibrium in the vertical direction and apply Equation (1)
∑F_y=n-mg
= mg (4)
We model the car as a particle in uniform circular motion in the horizontal direction. The force that enables the car to remain in its circular path is the force of static friction at the point of contact between road and tires. Apply Equation (2) to the horizontal direction:
∑F_x=f_s
=m*v^2/R
Substitute for f_s from Equation (3):
u_s=m*v^2/R
Substitute for n from Equation (4):
u_s*mg=m*v^2/R
u_s*g = v^2/R
Solve for u_s:
u_s= v^2/Rg (5)
Substitute numerical values:
u_s=0.375
(b)
The new coefficient of static friction between the tires and the pavement is:
u_s'=u_s/3
Substitute u_s/3 for u_s in Equation (5):
u_s/3=v^2/Rg
Solve for v:
v=14.4 m/s
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that initially the grass hopper is at rest at the ground position
Now the acceleration is given as
distance of the legs that it stretched is given as
so we have
Part b)
time taken to reach this speed is given as
Part c)
as the grass hopper reach the maximum height its final speed would be zero
so we will have