Question

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 begins 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.

Answer 1

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

Answer 2

The expression for the sum of the forces in the x-direction for the truck while braking $m(\mu_k g - a) = 0$

The given parameters;

• velocity of the semi, v = 26 m/s
• mass of the truck, = m
• coefficient of friction, μk = 0.26

The sum of the forces in the x-direction for the truck while braking is calculated as follows;

$\Sigma F_x = 0$

$F_k = ma\\\\\mu_k mg - ma =0\\\\m(\mu_k g - a) = 0$

Thus, the expression for the sum of the forces in the x-direction for the truck while braking $m(\mu_k g - a) = 0$

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