in the problem description it is F dog.for B it is (i.e. Fg = mg)for C it is (i.e. Fg = mg) # Understanding Forces on Box-Dog## Scenario DescriptionImagine taking your dog, Box-Dog, for a walk. Box-Dog moves with a force \( F_{\text{dog}} \) up a hill, which has a friction coefficient of \(\mu\). The hill is inclined at an angle \(\theta\), and the leash makes an angle \(\phi\) with respect to the normal. Box-Dog has a mass of \( m \).You can consider Box-Dog as a box moving with zero acceleration (\( a = 0 \, \text{m/s}^2 \)) for this analysis.### Task Breakdown#### a) Draw and Label a Free-Body Diagram for Box-Dog- **Diagram Explanation:** The diagram should represent Box-Dog as a box on an inclined plane.- **Forces to Label:** - **Gravitational Force (\(mg\))**: Acts vertically downward. - **Normal Force (\(N\))**: Acts perpendicular to the inclined surface. - **Frictional Force**: Acts parallel to the inclined surface, opposing the motion. - **Force \( F_{\text{dog}} \)**: Represents the force due to the leash, acting at an angle \(\phi\). #### b) Newton’s 2nd Law in the x-direction- **Objective:** Write the equation for forces in the horizontal direction along the plane.- **Equation Form:** Present the formula using variables and simplify.- **Considerations:** Reflect the forces using expressions like \( F=mg \) where appropriate.#### c) Newton’s 2nd Law in the y-direction- **Objective:** Write the equation for forces in the vertical direction.- **Equation Form:** Present the formula using variables and simplify.- **Considerations:** Reflect the forces using expressions like \( F=mg \) where appropriate.This exercise is designed to reinforce your understanding of the forces acting on an object on an inclined plane, combining concepts of friction, gravitational force, and inclined plane dynamics.

a) Let us assume the tension in the lease is T. Here f denotes the frictional force FBD: b) Applying Newton's 2nd law along x axis: Fdog-Tsinϕ-mgsinθ-f= maFdog-Tsinϕ-mgsinθ-f= m(0)Fdog-Tsinϕ-mgsinθ-μN=0 ....(1) c) Applying Newton's 2nd law along y axis: N+Tcosϕ=mgcosθN=mgcosθ-Tcosϕ ...(2) Applying (2) in (1) Fdog-Tsinϕ-mgsinθ-μ(mgcosθ-Tcosϕ)=0Fdog=Tsinϕ+mgsinθ+μ(mgcosθ-Tcosϕ)Fdog=Tsinϕ+mgsinθ+μmgcosθ-μTcosϕFdog=T(sinϕ-μcosϕ)+mg(sinθ+μcosθ)(answer of part a)