*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to
p of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P=300 N, the post slips at both B and C simultaneously.
1 answer:
Answer:
μ₁ = 0.1048
μ₂ = 0.1375
Explanation:
Using static equation can find in both point the moment and the forces so:
∑ M = F *d , ∑ F = 0
∑ M A = 0
N₁ * 3 - 200 * 9.81 * 1.5 = 0
N₁ = 981
∑ M y = 0
N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0
N₂ = 997.2 N
∑ M C = 0
F₁ * 1.75 - 300 * ⁴/₅ * 0.75 = 0
F₁ = 102.86
∑ M B = 0
300 * ⁴/₅ * 1 - F₂ * 1.75 = 0
F₂ = 137.14 N
The Force F1 and F2 related the coefficients of static friction
F₁ = μ₁ * N₁ ⇒ 102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048
F₂= μ₂ * N₂ ⇒ 137.14 N = μ₂ * 997 ⇒ μ₂ = 0.1375
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A) The velocity of sphere A after the collision is 1.00 m/s to the right
B) The collision is elastic
C) The velocity of sphere C is 2.68 m/s at a direction of
D) The impulse exerted on C is 4.29 kg m/s at a direction of
E) The collision is inelastic
F) The velocity of the center of mass of the system is 4.00 m/s to the right
Explanation:
A)
We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:
is the mass of sphere A
is the initial velocity of the sphere A (taking the right as positive direction)
is the final velocity of sphere A
is the mass of sphere B
is the initial velocity of the sphere B
is the final velocity of the sphere B
Solving for vA:
The sign is positive, so the direction is to the right.
B)
To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.
Before the collision:
After the collision:
The total kinetic energy is conserved: therefore, the collision is elastic.
C)
Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.
Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:
While the total momentum along the y-axis is zero:
We can now write the equations of conservation of momentum along the two directions as follows:
(1)
We also know the components of the momentum of B after the collision:
So substituting into (1), we find the components of the momentum of C after the collision:
So the magnitude of the momentum of C is
Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:
And the direction is
D)
The impulse imparted by B to C is equal to the change in momentum of C.
The initial momentum of C is zero, since it was at rest:
While the final momentum is:
So the magnitude of the impulse exerted on C is
And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore .
E)
To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.
The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:
The total kinetic energy after the collision is the sum of the kinetic energies of B and C:
Since the total kinetic energy is not conserved, the collision is inelastic.
F)
Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:
Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.
Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.
Learn more about momentum and collisions:
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