Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con
1 answer:
Answer:
Explanation:
When the block is displaced by x units
F= spring force
two springs are connected parallel
Writing Newtons second law, F = ma
a= x" ( differentiating x w.r.t time twice)
this the standard form of equation of oscillation spring mass system
This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration
since, Period
therefore,
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Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - = 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
The required torque at the axle, is given by the difference between the
moments of the applied forces.
The torque required is <u>19.62 N·m counterclockwise</u>
Reasons:
The given parameters are;
Mass of the disk, m = 5.0 kg
Location of the axle = Half the radius of the disk
Diameter of the disk, D = 40 cm = 0.4 m
Applied mass, 0.1 m from the axle = 15 kg
Applied mass, 0.3 m from the axle = 10 kg
Required:
Magnitude of torque at the axle that prevent the disk from rotating
Solution:
Torque needed = Clockwise moment - Counterclockwise moment
Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m
Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m
τ + Counterclockwise moment = Clockwise moment
τ + 14.715 N·m = 34.335 N·m
Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m
Torque required, τ = <u>19.62 N·m counterclockwise</u>
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<em>The probable question drawing obtained from a similar question online is attached</em>
Answer:
Explanation:
Given:
- mass of solid disk,
- radius of disk,
- force of push applied to disk,
- distance of application of force from the center,
<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>
where:
= static frictional force