A box of mass 20 kg is in the bed of a truck on a hill that is inclined at an angle of 24.1° above the horizontal. The tailgate of the truck is open. The truck is currently accelerating up the hill with a magnitude of 1.7 m/s2. themagnitude of the friction force from the truck on the box in this situation is 114.1N.
If the coeffcient of static friction between the box and the truck is 0.78, then what is the maximum acceleration that the truck can have if the box is not to slip
component of weight down truck bed = m g sin 24.1= 20*9.81*sin24.1
= 80.1 N
friction force up truck bed = 114.1 N
component of acceleration up truck bed= a cos 24.1
so
-80.1 + 114.1 = m a = 20 cos 24.1 * a
a = 34 / 20 cos 24.1
To find the maximum acceleration that the truck can have without causing the box to slip, we need to compare the gravitational force acting on the box with the maximum friction force that can be exerted by the truck on the box.
The gravitational force acting on the box can be calculated using the formula:
F_gravity = m * g
Where:
m = mass of the box (20 kg)
g = acceleration due to gravity (9.8 m/s^2)
F_gravity = 20 kg * 9.8 m/s^2 = 196 N
The maximum friction force that can be exerted by the truck on the box can be calculated using the formula:
F_friction = μ * N
Where:
μ = coefficient of static friction (0.78)
N = normal force
The normal force can be calculated by decomposing the weight force into its components parallel and perpendicular to the incline:
N = m * g * cos(θ)
Where:
θ = angle of inclination (24.1°)
N = 20 kg * 9.8 m/s^2 * cos(24.1°) ≈ 176.66 N
Finally, we can calculate the maximum friction force:
F_friction = 0.78 * 176.66 N ≈ 137.73 N
Now, to find the maximum acceleration, we equate the maximum friction force to the product of the mass of the box and the acceleration:
F_friction = m * a
where:
a = maximum acceleration (unknown)
Rearranging the equation, we can solve for the maximum acceleration:
a = F_friction / m
a = 137.73 N / 20 kg ≈ 6.89 m/s^2
Therefore, the maximum acceleration the truck can have without causing the box to slip is approximately 6.89 m/s^2.
To find the maximum acceleration that the truck can have without causing the box to slip, we can use the equation:
\(f_{\text{friction}} = \mu_s \cdot N\)
Where:
\(f_{\text{friction}}\) is the frictional force between the box and the truck,
\(\mu_s\) is the coefficient of static friction,
\(N\) is the normal force acting on the box.
In this case, the normal force can be calculated by:
\(N = m \cdot g \cdot \cos(\theta)\)
Where:
\(m\) is the mass of the box (20 kg),
\(g\) is the acceleration due to gravity (9.8 m/s^2),
\(\theta\) is the angle of the hill (24.1°).
Let's calculate the normal force first:
\(N = (20 \text{ kg}) \cdot (9.8 \text{ m/s}^2) \cdot \cos(24.1°)\)
Simplifying,
\(N \approx 174.09 \text{ N}\)
Now we can calculate the maximum frictional force before the box starts to slip:
\(f_{\text{friction}} = (0.78) \cdot (174.09 \text{ N})\)
Simplifying,
\(f_{\text{friction}} \approx 135.65 \text{ N}\)
Therefore, the maximum acceleration the truck can have without causing the box to slip is given by:
\(f_{\text{friction}} = m \cdot a\)
Substituting the values,
\(135.65 = (20 \text{ kg}) \cdot a\)
Simplifying,
\(a \approx 6.783 \text{ m/s}^2\)
Hence, the maximum acceleration the truck can have without causing the box to slip is approximately 6.783 m/s^2.