91Ó°ÊÓ

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is \(0.37\). (a) What is the frictional force on the crate from the floor? (b) What is the maximum magnitude \(f_{\max }^{\text {stat }}\) of the static frictional force under the circumstances? (c) Does the crate move? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Step by step solution

01

Calculate the Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. For a crate of mass \(m = 35 \text{ kg}\), the normal force \(F_n\) is given by \(F_n = mg\), where \(g = 9.8 \text{ m/s}^2\).

02

Calculate the Normal Force Value

Substitute the values: \[ F_n = 35 \text{ kg} \times 9.8 \text{ m/s}^2 = 343 \text{ N} \]

03

Calculate the Maximum Static Friction

The maximum static friction force \(f_{\text{max}}^{\text{stat}}\) is found using the coefficient of static friction \(\mu_s = 0.37\) and the normal force \(F_n\). The formula is \[ f_{\text{max}}^{\text{stat}} = \mu_s \times F_n = 0.37 \times 343 \text{ N} \]

04

Compute the Maximum Static Friction Value

Calculating the value: \[ f_{\text{max}}^{\text{stat}} = 0.37 \times 343 \text{ N} = 126.91 \text{ N} \]

05

Compare Applied Force to Static Friction

The applied force is 110 N. Determine the frictional force by comparing the applied force with the maximum static friction \(f_{\text{max}}^{\text{stat}}\). Since the applied force (110 N) is less than \(126.91 \text{ N}\), the static friction force equals the applied force: \[ f_{\text{friction}} = 110 \text{ N} \]

06

Determine Whether the Crate Moves

Because the applied force (110 N) is less than the maximum static friction \(f_{\text{max}}^{\text{stat}} = 126.91 \text{ N}\), the crate does not move.

07

Calculate the Least Vertical Pull to Move the Crate

Let the vertical pull be \(F_v\). The new normal force is \(F_n = mg - F_v\). The condition for moving the crate is \(f_{\text{max}}^{\text{stat}} = 110 \text{ N}\). Use \( \mu_s F_n = 110 \text{ N}\) to determine \(F_v\): \( 0.37(343 - F_v) = 110 \)

08

Solve for Vertical Pull

Rearrange to solve for \(F_v\): \[ 0.37 \times 343 - 0.37 \times F_v = 110 \ \Rightarrow 126.91 - 110 = 0.37 F_v \ \Rightarrow F_v = 46.95 \text{ N}\]

09

Calculate the Least Horizontal Pull

If a second worker adds a horizontal force \(F_h\), then the total horizontal force needed to overcome static friction is \( 110 \text{ N}\) plus the additional horizontal force. Hence, the least horizontal pull \(F_h\) must satisfy: \[ 110 \text{ N} + F_h = 126.91 \text{ N} \]

10

Solve for Horizontal Pull

Thus, \( F_h = 126.91 \text{ N} - 110 \text{ N} = 16.91 \text{ N}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force

The normal force is a supportive force applied by a surface to counterbalance the weight of an object resting on it. This force is always perpendicular to the surface in contact. For a crate with a mass of 35 kg, we use the formula:
\( F_n = mg \)
Here, \( F_n \) is the normal force, \( m \) is the mass, and \( g \) is the acceleration due to gravity, which is approximately 9.8 m/s². Plugging in the values:
\[ F_n = 35 \text{ kg} \times 9.8 \text{ m/s}^2 = 343 \text{ N} \]
This means the normal force exerted by the floor on the crate is 343 N.

Coefficient of Static Friction

The coefficient of static friction (\( \mu_s \)) is a dimensionless value that represents the frictional force between two static surfaces. It's a measure of how difficult it is to start moving one object over another. For this crate, the coefficient of static friction is given as 0.37.
To find the maximum static friction force (\( f_{\text{max}}^{\text{stat}} \)), use:
\[ f_{\text{max}}^{\text{stat}} = \mu_s \times F_n \]
By substituting,
\( f_{\text{max}}^{\text{stat}} = 0.37 \times 343 \text{ N} = 126.91 \text{ N} \)
This value indicates the maximum friction force before the crate starts moving.

Applied Force

The applied force refers to the force exerted by an external source, in this case, a worker pushing the crate horizontally with a force of 110 N. To check if the crate will move, compare this force with the maximum static friction force calculated earlier.
Since the applied force (110 N) is less than the maximum static friction force (126.91 N), the crate does not move. Thus, the static friction force equals the applied force of 110 N.
\( f_{\text{friction}} = 110 \text{ N} \)

Frictional Force

Frictional force is the resistance force that acts against the relative motion or tendency of such motion of two surfaces in contact. The static frictional force keeps the crate stationary because it adjusts to be equal and opposite to the applied force until it reaches its maximum limit.
The frictional force in this scenario is calculated as:
\( f_{\text{friction}} = 110 \text{ N} \).
This is because the crate doesn't move, so the static friction equals the applied force.

Horizontal Pull

When considering a horizontal pull by a second worker, the added force combined with the original push by the first worker must exceed the maximum static friction to move the crate.
The least additional horizontal pull (\( F_h \)) required to move the crate is found by:
\[ 110 \text{ N} + F_h = 126.91 \text{ N} \]
Solving for \( F_h \):
\( F_h = 126.91 \text{ N} - 110 \text{ N} = 16.91 \text{ N} \)
Therefore, at least 16.91 N of additional horizontal pull is necessary for the crate to start moving.