Questions: An object of mass m=5.21 kg is on the θ=32.5° incline plane with unknown applied force FA. Static friction coefficient of plane is μs=0.395 and kinetic friction coefficient of the plane μk=0.316. a) Consider object moving up with an acceleration of a=1.2 m / s^2. II. Find an algebraic equation for applied force and its numerical value?

Solution Steps

The forces acting on the object are:

1. Gravitational force (\( F_g \))
2. Normal force (\( F_N \))
3. Applied force (\( F_A \))
4. Kinetic friction force (\( F_k \))

The gravitational force can be resolved into two components:

• Parallel to the incline: \( F_{g,\parallel} = mg \sin \theta \)
• Perpendicular to the incline: \( F_{g,\perp} = mg \cos \theta \)

Given:

• \( m = 5.21 \, \text{kg} \)
• \( g = 9.81 \, \text{m/s}^2 \)
• \( \theta = 32.5^\circ \)

\[ F_{g,\parallel} = 5.21 \times 9.81 \times \sin(32.5^\circ) \approx 27.999 \, \text{N} \] \[ F_{g,\perp} = 5.21 \times 9.81 \times \cos(32.5^\circ) \approx 43.145 \, \text{N} \]

The normal force (\( F_N \)) is equal to the perpendicular component of the gravitational force: \[ F_N = F_{g,\perp} \approx 43.145 \, \text{N} \]

The kinetic friction force (\( F_k \)) is given by: \[ F_k = \mu_k F_N = 0.316 \times 43.145 \approx 13.634 \, \text{N} \]

Using Newton's second law along the incline: \[ F_A - F_{g,\parallel} - F_k = ma \]

Substitute the known values: \[ F_A - 27.999 - 13.634 = 5.21 \times 1.2 \] \[ F_A - 41.633 = 6.252 \] \[ F_A = 6.252 + 41.633 \approx 47.885 \, \text{N} \]

Final Answer