Respuesta :
Answer:8535 N
Explanation:
Given
mass of truck[tex](m_t)[/tex]=3600 kg
Truck is parked on a [tex]14 ^{\circ}[/tex] slope
so its sin component of weight will try to pull it down the slope and friction is resist its motion
Weight sin component[tex]=m_t\cdot gsin\theta =3600\times 9.8\times sin(14)[/tex]
=8535 N
Maximum Frictional force[tex]=\mu N[/tex]
[tex]N=mg\cos \theta [/tex]
[tex]f_r=0.9\times 3600\times 9.8\times \cos (14)=30,808.829 N[/tex]
so friction Force adjust itself to imparts 8535 N frictional force.
Friction force is the product of the coefficient of friction and the normal force. The friction force on the truck is 7,689.3423 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 3600\times 9.8 \times sin(14^o)\\\\f_r = 7,689.3423 \ N[/tex]
Hence, the friction force on the truck is 7,689.3423 N.
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