Solve Physics Problem: Coefficient of Friction on Steep Incline with Supertruck

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The problem involves calculating the coefficient of friction for a supertruck ascending a 65-degree incline with a specified acceleration and engine force. The initial calculations led to a coefficient of friction of approximately 0.80, but further analysis revealed a more accurate value of 0.82. The net force was determined by resolving the forces acting on the truck, including gravitational components and the applied engine force. The calculations confirmed that the friction force was 8054 N, leading to the final coefficient of friction being derived from the normal force. The discussion emphasizes the importance of correctly resolving forces and applying Newton's laws to find the solution.
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A supertruck is fighting its way up an steep 65 degree incline with an acceleration of 3.3 m/s^2. If the truck has a mass=1.00 tonne and an engine applied force of 1.55 X 10^4 Newtons, find the co-efficient of friction.

1 tonne = 1000 kg

**I got 0.80 as my answer...can anyone confirm this?
 
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Find the magnitude of gravity down the slope
Find the ideal acceleration of the truck by dividing force and mass.
Find a net force from the above two forces.
Find a net acceleration from this net force. (Divide by mass)
Compare this net acceleration to the acceleration provided.
Find the friction force by subtracting the two forces.
Divide by the normal force to get coefficient of friction.
Normal force is mgcos(\vartheta)
 
whozum said:
Find the magnitude of gravity down the slope
Find the ideal acceleration of the truck by dividing force and mass.
Find a net force from the above two forces.
Find a net acceleration from this net force. (Divide by mass)
Compare this net acceleration to the acceleration provided.
Find the friction force by subtracting the two forces.
Divide by the normal force to get coefficient of friction.
Normal force is mgcos(\vartheta)

Thank you...is 0.801 the answer?
 
Resultant force = ma

let friction = F.

so, resolving forces parallel to the slope, using g=9.81N/kg:
F = ma
1.55*10^4 - F - 1000gcos65 = 1000*3.3
F= 8054 N

F=uR, where u = coefficient of friction
u = F/R
u = 8054/(1000*9.81)
u = 0.82
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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