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:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
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The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:
where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:
where the negative sign indicates that the direction is backward.
Learn more about impulse:
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