An untethered block sits on a flatbed truck as it accelerates up an incline that makes an angle of 15° with respect to the horizontal. If the truck speeds up at a rate less than 2.69 m/s2, the block remains on the truck. If the truck speeds up at a rate equal to or greater than this value, however, the block slides off the truck. What is the coefficient of static friction between the truck and the block?

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ConcepcionPetillo ConcepcionPetillo · Answer 1 · 2019-09-25T08:37:23+02:00

Answer:

The coefficient of static friction is 0.524

Solution:

As per the question:

Angle made with the horizontal, [tex]\theta = 15^{\circ}[/tex]

Acceleration of the truck, a = 2.69[tex]m/s^{2}[/tex]

Now,

Forces are balanced along the x- direction:

When the block sits at rest and is about to slide:

[tex]F_{s} - mgsin\theta = 0[/tex] (1)

Also, static force of friction, [tex]F_{s}[/tex]:

[tex]F_{s} = \mu_{s}N[/tex]

where

N = mgcos[tex]\theta[/tex]

Thus

[tex]F_{s} = \mu_{s}mgcos\theta[/tex] (2)

Now, from eqn (1) and eqn (2):

[tex]\mu_{s}mgcos\theta - mgsin\theta = 0[/tex]

[tex]\mu_{s} = tan\theta[/tex]

Also,

a = [tex]\mu_{s}gcos\theta - gsin\theta[/tex]

[tex]\frac{2.69}{9.8} = \mu_{s}cos15^{\circ} - sin15^{\circ}[/tex]

[tex]0.2745 + 0.2588 = \mu_{s}cos15^{\circ}[/tex]

[tex]0.506 = \mu_{s}cos15^{\circ}[/tex]

[tex]\mu_{s} = 0.524[/tex]