A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are ????s = 0.9 and ????k = 0.5. The truck starts off up a 20° slope. What is the maximum acceleration the truck can have without the crate slipping out the back?

Answer :

erolkayacan

Answer:

Maximum acceleration of the truck is 5.25 m/s^2

Explanation:

To find the maximum acceleration the truck first we need to calculate friction between truck and wood. Because of wood is not moving, coefficient of c_{s} need to be used for it. Then the formula will be:

[tex]F_{s}=c_{s}*200*9.8*cos(20)\\F_{s}=0.9*200*9.8*0.94\\F_{s}=1657.62[/tex]

1657.62 Newton is the friction.

So force against friction need to be at most 1657.62 N. Then equation will be:

[tex]F_{s}=F_{a}+F_{m}\\1657.62=200*9.8*sin(20)+200*a*cos(20)\\1657.62=670.36+187.94*a\\987.26=187.94*a\\a=5.25[/tex]

Maximum acceleration of the truck need to be 5.25 m/s^2

5.25 m/s^2 is the Maximum acceleration of the truck.

How to Calculation of maximum acceleration?

Then We find the maximum acceleration to the truck first then we need to calculate friction between the truck and also wood. Because wood is not moving, the coefficient of c_{s} needs to be used for it. Then the formula that will be applicable is:

[tex]Fs = Сs * 200 * 9.8 * соs (20)[/tex]

[tex]Fs = 0.9 * 200 * 9.8 * 0.94[/tex]

[tex]Fs = 1657.62[/tex]

1657.62 Newton is the friction.

So, When the force against friction needs to be at most 1657.62 N. Then equation will be:

[tex]Fs = Fa + Fm[/tex]

[tex]1657.62 = 200 * 9.8 * sin (20) + 200* a * cos(20)[/tex]

[tex]1657.62 = 670.36 + 187.94 * a[/tex]

[tex]987.26 = 187.94 * a[/tex]

[tex]a = 5.25[/tex]

Thus, The Maximum acceleration of the truck needs to be 5.25 m/s^2.

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