Answer:
The positions are 0.0194 m and - 0.0194 m.
Explanation:
Given;
amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m
speed of simple harmonic motion is given as;
the maximum speed of the simple harmonic motion is given as;
when the speed equal one fourth of its maximum speed
Thus, the positions are 0.0194 m and - 0.0194 m.
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
Now;
Also,
where (angular velocity) =
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is
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
