A crate of oranges weighing \(180 \mathrm{N}\) rests on a flatbed truck $2.0 \mathrm{m}$ from the back of the truck. The coefficients of friction between the crate and the bed are \(\mu_{\mathrm{s}}=0.30\) and $\mu_{\mathrm{k}}=0.20 .\( The truck drives on a straight, level highway at a constant \)8.0 \mathrm{m} / \mathrm{s} .$ (a) What is the force of friction acting on the crate? (b) If the truck speeds up with an acceleration of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) what is the force of the friction on the crate? (c) What is the maximum acceleration the truck can have without the crate starting to slide?

We are given the following information: - Weight of crate, \(W = 180 \mathrm{N}\) - Distance from the back of the truck, \(d = 2.0 \mathrm{m}\) - Coefficients of friction: \(\mu_\mathrm{s} = 0.30\) and \(\mu_\mathrm{k} = 0.20\) - Truck's initial velocity, \(v = 8.0 \mathrm{m/s}\), and acceleration, \(a = 1.0 \mathrm{m/s^2}\) We also know that the normal force acting on the crate equals its weight.

To find the force of static friction acting on the crate, we use the formula \(f_\mathrm{s} \le \mu_\mathrm{s} \times N\), where \(f_\mathrm{s}\) is the static friction force and \(N\) is the normal force. Since the crate is in equilibrium (not accelerating), the static friction force (\(f_s\)) is equal to the weight of the crate. We know that the normal force equals the weight, \(N = W = 180\,\mathrm{N}\). Therefore: \(f_\mathrm{s} \le \mu_\mathrm{s} \times N = 0.30 \times 180 \mathrm{N} = 54 \mathrm{N}\) In this case, \(f_s = W = 54\, \mathrm{N}\).

When the truck accelerates, the frictional force changes. To calculate the new force, we use Newton's second law along the horizontal axis: \(f_\mathrm{friction} = ma_\mathrm{crate}\). Since the crate is still not sliding, the static friction force is acting: \(f_\mathrm{friction} = m \times a_\mathrm{truck}\) We can find the mass of the crate using the weight: \(m = W / g = 180\,\mathrm{N} / 9.81\,\mathrm{m/s^2} \approx 18.34\,\mathrm{kg}\). Now, we find the frictional force: \(f_\mathrm{friction} = 18.34 \,\mathrm{kg} \times 1.0 \,\mathrm{m/s^2} = 18.34 \,\mathrm{N}\) Since this frictional force is less than the maximum static friction force, the crate won't start sliding.

We now find the maximum acceleration of the truck before the crate starts to slide. We use the maximum static friction force and Newton's second law to find the acceleration: \(f_\mathrm{s, max} = m \times a_\mathrm{truck, max}\) \(a_\mathrm{truck, max} = \frac{f_\mathrm{s, max}}{m} = \frac{54\,\mathrm{N}}{18.34\,\mathrm{kg}} \approx 2.94\,\mathrm{m/s^2}\) So, the maximum acceleration the truck can have without the crate starting to slide is about \(2.94\,\mathrm{m/s^2}\). In summary: (a) The force of friction acting on the crate at rest is \(54\,\mathrm{N}\). (b) If the truck accelerates at \(1.0\,\mathrm{m/s^2}\), the force of friction on the crate is \(18.34\,\mathrm{N}\). (c) The maximum acceleration the truck can have without the crate starting to slide is approximately \(2.94\,\mathrm{m/s^2}\).