Question

a worker wants to turn over a uniform 1230 nn rectangular crate by pulling at 53.0?? on one of its vertical sides (figure 1). the floor is rough enough to prevent the crate from slipping.

Answer 1

To turn over the rectangular crate without slipping, the worker needs to exert a force greater than or equal to 1230 N parallel to the vertical side at an angle greater than 53.0 degrees.

Step-by-step explanation:

To turn over the crate without it slipping, the worker needs to apply a force greater than the maximum static friction force to overcome the resistance to tipping. The maximum static friction force is calculated as \
`(f_{\text{friction}} = \mu_s \cdot N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.`

The normal force exerted by the worker on the vertical side of the crate is \(N = mg\), where \(m\) is the mass of the crate and \(g\) is the acceleration due to gravity. Given that the crate is uniform, the mass can be converted to weight (\(mg\)).

However, the worker's force at an angle of 53.0 degrees might not necessarily overcome the static friction and initiate the crate's tipping. Therefore, the worker should exert a force greater than the calculated maximum static friction force
`(\(f_{\text{friction}}\))`to ensure the crate's successful turnover without slipping.

Understanding the principles of static friction, forces, and torque is crucial in situations where objects need to be manipulated or moved without slipping or sliding.

Answer 2

The worker must apply a force of at least 982.18 Newtons to turn over the rectangular crate by pulling at an angle of 53.0 degrees on one of its vertical sides.

To turn over the rectangular crate by pulling at an angle of 53.0 degrees on one of its vertical sides, you can follow these steps:

1. Identify the forces involved:

- The force applied by the worker (F_applied).

- The weight of the crate (W) acting downward.

2. Break down the weight (W) into its components:

- Wx: The horizontal component of the weight.

- Wy: The vertical component of the weight.

3. Calculate the weight components:

- Wx = W * sin(53.0°)

- Wy = W * cos(53.0°)

4. To tip the crate over, the worker must apply a force greater than or equal to the horizontal component (Wx) of the weight. So, F_applied ≥ Wx.

5. Calculate F_applied:

- sin(53.0°)

6. Plug in the values and calculate:

- F_applied ≥ W * sin(53.0°)

- F_applied ≥ 1230 N * sin(53.0°)

- F_applied ≥ 1230 N * 0.7986

- F_applied ≥ 982.18 N

So, the worker must apply a force of at least 982.18 Newtons to turn over the rectangular crate by pulling at an angle of 53.0 degrees on one of its vertical sides.