1) 
We can solve this part by using Newton's second law:
(1)
where
F is the net force
m is the mass
a is the acceleration
There are two forces acting on the boat:
forward
backward
So the net force is
We know that the mass of the boat is
m = 1177.5 kg
So we can now use eq.(1) to find the acceleration:
2) 161.0 m
We can solve this part by using the following suvat equation:
where
s is the distance travelled
u is the initial velocity
t is the time
a is the acceleration
Here we have
u = 0 (the boat starts from rest)
Substituting t = 17.7 s, we find the distance covered:
3) 18.2 m/s
The speed of the boat can be found with the following suvat equation
where
v is the final velocity
u is the initial velocity
t is the time
a is the acceleration
In this case we have
u = 0 (the boat starts from rest)
And substituting t = 17.7 s, we find the final velocity:
And the speed is just the magnitude of the velocity, so 18.2 m/s.
Vx=cos60(4)
x-component of velocity
<span>If you think about it, it makes a right triangle when you combine all the different types of forces together such as v, vx and vy. Then, you can use trigonometry and soh cah toa in order to figure out vx. </span>
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
