### Problem Statement:#### Part (a):At a construction site, a pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown below. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666.**Question:**What is the weight of the heaviest pallet of bricks that can be supported this way? Start with appropriate free-body diagrams.#### Part (b):What is the friction force on the upper crate under the conditions of part (a)?### Diagram Description:The provided diagram depicts the following scenario:- A pallet of bricks is suspended over the edge of a building's roof via a pulley system.- The other end of the rope is attached horizontally to the lower of two stacked crates on the roof.- The weights of the two crates are indicated as 150 lb for the upper crate and 250 lb for the lower crate.**Visual Elements:**1. Two crates stacked on top of one another on the roof. The bottom crate, weighing 250 lb, is connected to a rope pulling horizontally.2. The top crate on the stack weighs 150 lb.3. The rope passes through a pulley system secured at the roof’s edge and supports a pallet of bricks, which hangs vertically.### Free-Body Diagram Analysis:1. **For the pallet of bricks:** - Weight (W_bricks) = the force due to gravity acting on the pallet. - Tension (T) = the tension in the rope due to the weight of the bricks.2. **For the lower crate:** - Normal force (N_lower) = the force exerted by the roof surface on the lower crate. - Weight (W_lower) = 250 lb. - Friction force (F_friction) = μ * N_lower, where μ = 0.666. - Tension force (T) = horizontally acting force pulling the lower crate.3. **For the upper crate:** - Normal force (N_upper) = the force exerted by the lower crate. - Weight (W_upper) = 150 lb.### Equations:1. **Static Friction Force: F_friction = μ * N_lower**.2. **For equilibrium of the lower crate:** T = F_friction

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Description: ### Problem Statement:#### Part (a):At a construction site, a pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown below. The rope pulls horizontal

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(a) At a construction site, a pallet of bricks is to be suspended by attach- ing a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown below. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666. (a) What is the weight of the heaviest pallet of bricks that can be supported this way? Start with appropriate free-body diagrams. (b) What is the friction force on the upper crate under the conditions of part (a)? 150 lb 250 lb Bricks

(a) At a construction site, a pallet of bricks is to be suspended by attach- ing a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown below. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666. (a) What is the weight of the heaviest pallet of bricks that can be supported this way? Start with appropriate free-body diagrams. (b) What is the friction force on the upper crate under the conditions of part (a)? 150 lb 250 lb Bricks

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Newton’s laws of motion describe the relationship between the motion of an object and forces acting on it. Newton’s law of motion has two types as follows:

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### Problem Statement:

#### Part (a):

At a construction site, a pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the roof of a building, as shown below. The rope pulls horizontally on the lower crate, and the coefficient of static friction between the lower crate and the roof is 0.666.

**Question:**
What is the weight of the heaviest pallet of bricks that can be supported this way? Start with appropriate free-body diagrams.

#### Part (b):

What is the friction force on the upper crate under the conditions of part (a)?

### Diagram Description:

The provided diagram depicts the following scenario:
- A pallet of bricks is suspended over the edge of a building's roof via a pulley system.
- The other end of the rope is attached horizontally to the lower of two stacked crates on the roof.
- The weights of the two crates are indicated as 150 lb for the upper crate and 250 lb for the lower crate.

**Visual Elements:**
1. Two crates stacked on top of one another on the roof. The bottom crate, weighing 250 lb, is connected to a rope pulling horizontally.
2. The top crate on the stack weighs 150 lb.
3. The rope passes through a pulley system secured at the roof’s edge and supports a pallet of bricks, which hangs vertically.

### Free-Body Diagram Analysis:

1. **For the pallet of bricks:**
- Weight (W_bricks) = the force due to gravity acting on the pallet.
- Tension (T) = the tension in the rope due to the weight of the bricks.

2. **For the lower crate:**
- Normal force (N_lower) = the force exerted by the roof surface on the lower crate.
- Weight (W_lower) = 250 lb.
- Friction force (F_friction) = μ * N_lower, where μ = 0.666.
- Tension force (T) = horizontally acting force pulling the lower crate.

3. **For the upper crate:**
- Normal force (N_upper) = the force exerted by the lower crate.
- Weight (W_upper) = 150 lb.

### Equations:

1. **Static Friction Force: F_friction = μ * N_lower**.
2. **For equilibrium of the lower crate:** T = F_friction

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