A crate of mass 52.8 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the trucks flatbed is 0.351, and the coefficient of kinetic friction is 0.305.
(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the trucks flatbed?
(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?
(a)
ma=F(fr) = μ(s) •N= μ(s) •m•g,
a= μ(s)•g,
(b)
ma1=F1(fr) = μ(k) •N= μ(k) •m•g,
a1= μ(k)•g,
A stone thrown vertically upward from the ground level with a speed of 25m/s. At what time will be 5m above the ground
To solve this problem, we will first determine the maximum acceleration of the truck, using the coefficient of static friction. Then, we will calculate the acceleration of the crate relative to the ground, once the truck starts moving and the crate starts sliding.
(a) To find the maximum acceleration of the truck so that the crate does not slide relative to the truck's flatbed, we need to compare the force of static friction and the maximum frictional force that can be exerted. The maximum frictional force is given by:
\(F_{\text{friction,max}} = \mu_s \cdot F_N\)
where:
\(F_{\text{friction,max}}\) is the maximum frictional force,
\(\mu_s\) is the coefficient of static friction,
\(F_N\) is the normal force between the crate and the truck's flatbed.
The normal force is equal to the weight of the crate, which can be calculated using the formula:
\(F_N = m \cdot g\)
where:
\(m\) is the mass of the crate,
\(g\) is the acceleration due to gravity.
Therefore, the maximum frictional force is:
\(F_{\text{friction,max}} = \mu_s \cdot m \cdot g\)
To prevent the crate from sliding, the maximum frictional force must be equal to the net force acting on the crate:
\(F_{\text{net}} = m \cdot a\)
where:
\(F_{\text{net}}\) is the net force on the crate,
\(a\) is the maximum acceleration of the truck.
Setting these two forces equal, we have:
\(m \cdot a = \mu_s \cdot m \cdot g\)
Simplifying the equation, we get:
\(a = \mu_s \cdot g\)
Substituting the given values, we get:
\(a = 0.351 \cdot 9.8\, \text{m/s}^2\)
\(a \approx 3.44\, \text{m/s}^2\)
Therefore, the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed is approximately \(3.44\, \text{m/s}^2\).
(b) Once the truck exceeds this maximum acceleration, the crate starts sliding along the truck's flatbed. Now, we need to find the acceleration of the crate relative to the ground.
The frictional force acting on the crate while sliding is the kinetic friction, given by:
\(F_{\text{friction,kinetic}} = \mu_k \cdot F_N\)
where:
\(F_{\text{friction,kinetic}}\) is the kinetic frictional force,
\(\mu_k\) is the coefficient of kinetic friction,
\(F_N\) is the normal force.
To calculate the net force on the crate, we need to subtract the frictional force from the force exerted by the truck:
\(F_{\text{net}} = F_{\text{truck}} - F_{\text{friction,kinetic}}\)
Where \(F_{\text{truck}}\) is the force exerted by the truck.
As the crate is in contact with the truck, it will experience the same acceleration as the truck. Therefore, we have:
\(a_{\text{crate}} = \frac{{F_{\text{net}}}}{{m}}\)
Substituting the values, we get:
\(a_{\text{crate}} = \frac{{F_{\text{truck}} - F_{\text{friction,kinetic}}}}{{m}}\)
To find \(F_{\text{truck}}\), we need to use Newton's second law:
\(F_{\text{truck}} = m \cdot a_{\text{truck}}\)
Substituting this into the previous equation, we have:
\(a_{\text{crate}} = \frac{{m \cdot a_{\text{truck}} - F_{\text{friction,kinetic}}}}{{m}}\)
Simplifying the equation, we get:
\(a_{\text{crate}} = a_{\text{truck}} - \frac{{F_{\text{friction,kinetic}}}}{{m}}\)
Substituting the given values, we get:
\(a_{\text{crate}} = a_{\text{truck}} - \frac{{0.305 \cdot m \cdot g}}{{m}}\)
\(a_{\text{crate}} = a_{\text{truck}} - 0.305 \cdot g\)
Therefore, the acceleration of the crate relative to the ground is \(a_{\text{crate}} = a_{\text{truck}} - 0.305 \cdot g\).