Respuesta :
Answer:
Explanation:
given
spring constant k = 880 N/m
mass m = 150 kg
Normal force will be equal to the component of weight of mass m which is perpendicular to the inclined surface
= mgcosθ
So normal force
FN = mgcosθ j , as it acts in out of plane direction .
b )
Fricrion force acting in upward direction = μs mgcosθ
component of weight acting in downward direction = mgsinθ
restoring force by spring on block in downward direction
= kd
= 880d
F( total ) = (μs mgcosθ - mgsinθ - 880d )i
c )
for balance
mgsinθ = kd
sinθ = kd / mg
= 880 x .1 / 150 x 9.8
= 88 / 1470
.0598
θ = 3.4 degree
d )
d = mgsinθ / k
150 x 9.8 sin45 / 880
= 1.18 m
The traslational equilibrium condition and Hooke's law allow us to find the results for the problem is static equilibrium are:
a) The expression for the normal force is: N = m g cos θ
b) The expression for the forces in the x-axis is:
kx + mg (sin θ - mg cos θ)=0
c) The angle of the plane is θ = 3.4º
d) For angle of 45º the compression of the spring is: x = 1.18 m
Newton's second law gives a relationship between the force, the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it reduces to the equilibrium condition.
Σ F = 0
Where bold letters indicate vectors, F is the external force
The reference system with respect to which to carry out the measurements and decomposition of the vector is indicated in the exercise, where the x-axis is parallel to the plane and is upwards is positive, the y=axis is perpendicular to the plane
The free-body diagram is a representation of the forces without the details of the bodies. In the attachment we can see a free-body diagram of this system.
Let's write the equilibrium condition for each exercise
a) y-axis
N- W_y = 0
N = [tex]W_y[/tex]
Let's use trigonometry to find the components of the weight
sin θ θ = [tex]\frac{W_x}{W}[/tex]
cos θ = [tex]\frac{W_y}{W}[/tex]
Wₓ = W sin θ
W_y = W cos θ
We substitute
N = m g cos θ
b) x-axis
Just before starting to move the acceleration is zero
[tex]F_e[/tex] - fr - Wₓ = 0
The friction force that is given by the expression
fr = μ N
We substitute
fr = μ mg cos θ
Hooke's law states that the force recovered from a spring is proportional to the displacement from equilibrium position
[tex]F_e[/tex] = - kx
We substitute
kx - μ mg cos θ - m g sin θ = 0 (1)
c) In this part they indicate that the friction force is zero and ask how much the angle of inclination is worth
kx - m g sin θ = 0
θ = sin⁻¹ [tex]\frac{kx}{mg}[/tex]
θ = sin⁻¹ ( [tex]\frac{880 \ 0.1}{150 \ 9.8}[/tex])
θ = 3.4º
d) How much the spring is compressed for an angle of 45º
We write equation 1
kx - m g sin θ = 0
x = [tex]\frac{mg}{k} \ sin \theta[/tex]
x = [tex]\frac{150 \ 9.8 }{880} \ sin 45[/tex]
x = 1.18 m
In conclusion, using the traslational equilibrium condition and Hooke's law we can find that the results for the problem is equilibrium are:
a) The expression for the normal force is: N = m g cos tea
b) The expression for the forces in the x- axis is:
kx + mg (sin θ - μ cos θ) = 0
c) The angle of the plane is θ = 3.4º
d) For angle θ = 45º the compression of the spring is: x = 1.18 m
Learn more here: brainly.com/question/13861141
