Carla holds a ball 1.5 m above the ground. Daniel, leaning out of a car window, also holds a ball 1.5 m above the ground. Daniel
2 answers:
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(a) 23.4
The fiber-to-matrix load ratio is given by
where
is the fiber elasticity module
is the matrix elasticity module
is the fraction of volume of the fiber
is the fraction of volume of the matrix
Substituting,
(1)
(b) 44,594 N
The longitudinal load is
F = 46500 N
And it is sum of the loads carried by the fiber phase and the matrix phase:
(2)
We can rewrite (1) as
And inserting this into (2):
Solving the equation, we find the actual load carried by the fiber phase:
(c) 1,906 N
Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:
(2)
Using
F = 46500 N
We can immediately find the actual load carried by the matrix phase:
(d) 437 MPa
The cross-sectional area of the fiber phase is
where
is the total cross-sectional area
Substituting , we have
And the magnitude of the stress on the fiber phase is
(e) 8.0 MPa
The cross-sectional area of the matrix phase is
where
is the total cross-sectional area
Substituting , we have
And the magnitude of the stress on the matrix phase is
(f)
The longitudinal modulus of elasticity is
While the total stress experienced by the composite is
So, the strain experienced by the composite is
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
