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Math  /  Algebra

Question22. (II) A flatbed truck is carrying a heavy crate. The coefficient of static friction between the crate and the bed of the truck is 0.75 . What is the maximum rate at which the driver can decelerate and still avoid having the crate slide against the cab of the truck?

Studdy Solution

STEP 1

1. The coefficient of static friction (μs \mu_s ) between the crate and the truck bed is 0.75.
2. We are trying to find the maximum deceleration rate of the truck.
3. The crate does not slide if the static frictional force is greater than or equal to the force due to deceleration.

STEP 2

1. Understand the forces acting on the crate.
2. Write the equation for static friction.
3. Relate static friction to deceleration.
4. Solve for maximum deceleration.

STEP 3

Understand the forces acting on the crate. The force of static friction (fs f_s ) is what prevents the crate from sliding. This force is given by:
fs=μs⋅N f_s = \mu_s \cdot N
where N N is the normal force. Since the crate is on a flat surface, N=mg N = mg , where m m is the mass of the crate and g g is the acceleration due to gravity.

STEP 4

Write the equation for static friction. Substitute the expression for the normal force into the static friction equation:
fs=μs⋅mg f_s = \mu_s \cdot mg

STEP 5

Relate static friction to deceleration. The maximum static frictional force is equal to the maximum force that can be applied to the crate without it sliding. This force is also equal to the mass of the crate times the maximum deceleration (amax a_{\text{max}} ):
fs=m⋅amax f_s = m \cdot a_{\text{max}}

STEP 6

Solve for maximum deceleration. Set the two expressions for the static frictional force equal to each other:
μs⋅mg=m⋅amax \mu_s \cdot mg = m \cdot a_{\text{max}}
Cancel the mass (m m ) from both sides:
μs⋅g=amax \mu_s \cdot g = a_{\text{max}}
Substitute the given value of μs=0.75 \mu_s = 0.75 and g=9.8 m/s2 g = 9.8 \, \text{m/s}^2 :
amax=0.75⋅9.8 a_{\text{max}} = 0.75 \cdot 9.8
Calculate the maximum deceleration:
amax=7.35 m/s2 a_{\text{max}} = 7.35 \, \text{m/s}^2
The maximum rate at which the driver can decelerate without the crate sliding is:
7.35 m/s2 \boxed{7.35 \, \text{m/s}^2}

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