Respuesta :
Answer:
fr = 10145 N
Explanation:
To solve this exercise let's use Newton's second law
F = ma
Let's write a reference system, where the x-axis is parallel to the slope and the y-axis is perpendicular
Let's decompose the weight that is not in the direction of the axes
cos θ = Wy / W
sin θ = Wx / W
Wy = W cos θ
Wx = W sin θ
We write Newton's second law on each axis
Case 1. Truck stopped on the upward slope
X axis
fr - Wx = 0
fr = Wx
YAxis
N- Wy = 0
N = Wy
The friction force has the equation
fr = μ N
Let's calculate the force of friction
fr = W sin θ
fr = mg sin θ
fr = 4000 9.8 without 15
fr = 10145 N
Case2 Truck stopped on the descending slope
X axis
Wx -fr = 0
fr = Wx
fr = W sin θ = m g sin 15
fr = 10145N
We can see that the value of the friction force is the same regardless of the direction of the slope
Friction force is the product of the coefficient of friction and the normal force. The friction force on the truck is 9,140.4534 N.
What is friction force?
The friction force is the product of the coefficient of friction and the normal force.
[tex]\rm f_r = \mu N[/tex]
Let's assume a reference system, where the x-axis is parallel to the slope and the y-axis is perpendicular.
If we decompose the weight force that is not in the direction of the axes, then the horizontal and the vertical weight force can be written as,
[tex]W_y = {\rm W\cdot cos (\theta)}\\\\Wx = {\rm W\cdot sin (\theta)}[/tex]
Now, according to Newton's second law of motion, the equilibrium of the forces can be written as
X-axis
[tex]\rm f_r - W_x = 0\\\\f_r = W_x[/tex]
Y-Axis
[tex]N- W_y = 0\\\\N = W_y[/tex]
We know that the friction force can be written as [tex]\rm f_r = \mu N[/tex], therefore,
[tex]\rm f_r = \mu N\\\\f_r = \mu \times W \cdot sin\rm \theta\\\\f_r = \mu \times mg \cdot sin \theta\\\\f_r =0.9 \times 4000\times 9.8 \times sin(15^o)\\\\f_r = 9,140.4534\ N[/tex]
Hence, the friction force on the truck is 9,140.4534 N.
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