An \(80.0-\mathrm{N}\) crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at \(20.0^{\circ}\) to the ground. (a) What is the normal force exerted on the crate by the ramp? (b) The interaction partner of this normal force has what magnitude and direction? It is exerted by what object on what object? Is it a contact or a long-range force? (c) What is the static frictional force exerted on the crate by the ramp? (d) What is the minimum possible value of the coefficient of static friction? (e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. Find the magnitude and direction of the contact force.

To find the normal force exerted on the crate by the ramp, we need to decompose the weight of the crate into components parallel and perpendicular to the ramp. The normal force has equal magnitude but opposite direction to the component of weight perpendicular to the ramp. Let's first find that component: 1. Calculate the angle between the weight and the perpendicular component, which equals the angle of inclination of the ramp (\(20.0^{\circ}\)). 2. Use the trigonometric functions to find the component of the weight perpendicular to the ramp: \(W_{\perp} = W \cdot \cos(\theta)\), where \(W\) is the weight of the crate, and \(\theta\) is the angle between the weight and the perpendicular component. 3. The normal force exerted on the crate by the ramp, \(N\), is equal in magnitude but opposite in direction to the component of weight perpendicular to the ramp: \(N = W_{\perp}\).

Newton's Third Law states that every action has an equal and opposite reaction. The interaction partner of the normal force in this case is exerted by the crate on the ramp. This force has equal magnitude but opposite direction to the normal force exerted on the crate by the ramp. It is a contact force since it acts at the point of contact between the crate and the ramp.

Since the crate is at rest, there is a static frictional force acting on it to counteract the component of weight parallel to the ramp. To find the static frictional force, we need to find the component of the weight parallel to the ramp and equate it to the friction force: 1. Use the trigonometric functions to find the component of the weight parallel to the ramp: \(W_{\parallel} = W \cdot \sin(\theta)\), where \(\theta\) is the angle between the weight and the perpendicular component. 2. According to the property of static friction, the maximum static frictional force is equal to the component of weight parallel to the ramp: \(f_s = W_{\parallel}\).

To find the minimum possible value of the coefficient of static friction, we'll use the equation \(f_s = \mu_s N\), where \(f_s\) is the static frictional force, \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force. Rearrange the equation to find the coefficient of static friction: \(\mu_s = \frac{f_s}{N}\).

The contact force exerted on the crate by the ramp is the vector sum of the normal force and the static frictional force. 1. Express the normal force and the static frictional force as vectors. 2. Add the two vectors to find the contact force vector. 3. Calculate the magnitude of the contact force vector using the Pythagorean theorem: \(F_c = \sqrt{N^2 + f_s^2}\). 4. Find the direction of the contact force vector. This will be the angle between the contact force vector and the normal force vector. Use the following equation to find the angle: \(\alpha = \arctan\left(\frac{f_s}{N}\right)\).