Let's break down the problem step by step. ### **Given:** - **Bar AB:** mass \(2m\), uniform, thin - **Disk D:** mass \(3m\), uniform, rigidly attached to bar AB - **System is released from rest on the corner of a table** - **Coefficient of static friction between bar and table:** \(\mu_s\) ### **To Find:** (a) **Normal force** between the body and the table as a function of \(m, g, \theta\). (b) **Friction force** between the body and the table as a function of \(m, g, \theta\). (c) **Angle \(\theta\) at which the body will start to slip relative to the table** (as a function of \(\mu_s\)). --- ## **Step 1: Free Body Diagram & System Weight** Let: - The *bar* is at an angle \(\theta\) with the horizontal. - All weight acts vertically downward. **Total mass:** \(M = 2m + 3m = 5m\) **Total weight:** \(W = 5mg\) --- ## **(a) Normal Force \(N\) as a function of \(m, g, \theta\)** The normal force is the **vertical reaction** from the table. If the table is horizontal and all weight is supported by the table (no other supports or vertical forces), the normal force **must balance the total weight**: \[ N = W = 5mg \] #### **But...** If the system is at an angle, and if there's any component of force along the bar (say if it's not flat), the normal force might change. But **unless the setup is such that some weight is not supported (e.g., part of the bar is off the table), the normal force remains \(5mg\)**. **If the bar is on the verge of slipping, the normal force is still \(N = 5mg\)**. **Answer:** \[ \boxed{N = 5mg} \] --- ## **(b) Friction Force \(F_f\) as a function of \(m, g, \theta\)** The friction force **prevents sliding**. The gravitational force creates a component along the bar that tries to slide it. ### **Assuming the bar makes an angle \(\theta\) with the horizontal:** - **Weight acts vertically down:** \(5mg\) - **Friction acts along the table surface, opposing motion.** #### **Decompose the weight:** If the bar is at angle \(\theta\), the **component of weight down the incline** is: \[ W_{\text{down along bar}} = 5mg \sin\theta \] Friction resists this motion: \[ F_f = 5mg \sin\theta \] **Answer:** \[ \boxed{F_f = 5mg \sin\theta} \] --- ## **(c) Angle \(\theta\) at which the body will start to slip** The **maximum static friction** is: \[ F_{f, \text{max}} = \mu_s N = \mu_s \cdot 5mg \] Slipping occurs when the **component of gravity along the bar equals the maximum static friction:** \[ 5mg \sin\theta = \mu_s \cdot 5mg \] \[ \sin\theta = \mu_s \] \[ \boxed{ \theta = \arcsin(\mu_s) } \] --- ## **Summary of Answers** **(a)** \(\boxed{N = 5mg}\) **(b)** \(\boxed{F_f = 5mg \sin\theta}\) **(c)** \(\boxed{\theta = \arcsin(\mu_s)}\) --- ### **Note** If the orientation or placement is more complex (e.g., the disk is hanging off the table, changing torques, etc.), please upload a figure for a more precise solution. The above assumes all mass is on the table, and the bar is inclined at \(\theta\) to the horizontal. If not, please clarify or provide the diagram for a tailored answer.