a truck is carrying a 2800 kg crate of heavy machinery. if the coefficient of static friction between the crate and the bed of the truck is 0.55, what is the maximum rate the driver can decelerate when coming to a stop in order to avoid crushing the cab with the crate?
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Friction is the force causing the crate to accelerate.
Forcefriction= masscrate*acceleration
mu(masscrate*g)=masscrate(acceleration
solve for acceleration
To solve for the acceleration, we can use the equation:
Force_friction = mass_crate * acceleration
Here, the force of friction is given by the product of the coefficient of static friction (μ) and the mass of the crate (m_crate), multiplied by the acceleration due to gravity (g). Equating this to the mass of the crate (m_crate) multiplied by the acceleration (a), we have:
μ * m_crate * g = m_crate * a
Now, we can cancel out the mass of the crate on both sides of the equation:
μ * g = a
The maximum acceleration the truck can have is when the force of friction is at its maximum, which occurs when the truck is on the verge of slipping. In this case, the maximum force of static friction is given by:
Friction_max = μ * m_crate * g
To avoid crushing the cab with the crate, the maximum acceleration must be less than or equal to the maximum force of static friction divided by the mass of the crate:
a_max = Friction_max / m_crate
Substituting the value of Friction_max derived earlier, we have:
a_max = (μ * m_crate * g) / m_crate
Simplifying the equation:
a_max = μ * g
Therefore, the maximum deceleration rate the driver can have to avoid crushing the cab with the crate is equal to the product of the coefficient of static friction (μ) and the acceleration due to gravity (g). It does not depend on the mass of the crate.