A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
s to slide. The truck has mass m and a coefficient of kinetic friction between the tires and the road of μk = 0.26.
Write an expression for the sum of the forces in the x-direction for the truck while braking.
Write an expression for the sum of the forces in the x-direction for the truck while braking.
1 answer:
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Refer to the provided diagram in the attachment
fx= fcosθ (fx: x-direction component of friction force; f: frictional force)
Fbx= Fbcosθ (Fbx: x-direction component of braking force; Fb: braking force)
Wx= Wtanθ (Wx: x-direction component of weight; W: weight of the semi)
Sum of forces in the x-direction = 0
fx + Fbx = Wx
fcosθ + Fbcosθ =Wtanθ
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Response:
The car's acceleration magnitude is 35.53 m/s²
Details:
Given;
acceleration of the truck, = 12.7 m/s²
mass of the truck, = 2490 kg
mass of the car, = 890 kg
let the acceleration of the car during the collision =
Using Newton's third law of motion;
The force exerted by the truck equals the force exerted by the car.
The car's force acts in the opposite direction.
Thus, the car's acceleration magnitude is 35.53 m/s²