Max Tension Force for Box Not to Slip: 24.5N

In summary, the problem involves a horizontal rope pulling a 10kg wood sled and a 5.0kg wood box on frictionless snow, with a coefficient of static friction of 0.5 for wood on wood. The question asks for the largest tension force that would prevent the box from slipping. Using the equation f(static) = mu * f(normal), the maximum force is calculated to be 24.5N. However, in order to accelerate both the box and the sled, the tension force in the rope must also account for the acceleration of the box, resulting in a different value.
  • #1
zellster87
15
0

Homework Statement



A horizontal rope pulls a 10kg wood sled across frictionless snow. A 5.0kg wood box rides on the sled. The coefficient of static friction for wood on wood is mu = 0.5.

What is the largest tension force for which the box doesn't slip?

Homework Equations



f(static) = mu * f(normal)


The Attempt at a Solution



I don't really see how the mass of the sled is relevant. But here's what I did:

f(static) = 0.5 * 5(9.81) = 24.5N
 
Physics news on Phys.org
  • #2
You've calculated the maximum horizontal force the sled can exert on the 5 Kg box. That is not the whole answer to the question though.

In order to exert that force and accelerate the box, the sled itself must also be accelerating, and therefore you must calculate the acceleration that the 5 Kg box would undergo with a force of 24.5 N acting on it. The max tension in the rope would be the force required to accelerate the box AND the sled.
 
  • #3
ah ok. Thanks for the reply.
 

Related to Max Tension Force for Box Not to Slip: 24.5N

1. What is the purpose of determining the max tension force for a box?

The purpose of determining the max tension force for a box is to ensure that the box does not slip or move when it is being lifted or pulled. This is important for safety and stability during transportation or handling of the box.

2. How is the max tension force for a box calculated?

The max tension force for a box is calculated by using the formula F = μN, where F is the force applied, μ is the coefficient of friction between the box and the surface it is resting on, and N is the normal force exerted on the box by the surface. The value of μ is usually provided by the manufacturer or can be determined through experiments.

3. What factors can affect the max tension force for a box?

The max tension force for a box can be affected by several factors such as the weight of the box, the surface it is resting on, the coefficient of friction between the box and the surface, and the angle at which the force is applied. Additionally, any external forces acting on the box, such as wind or vibrations, can also affect the max tension force.

4. How does the max tension force for a box impact its handling and transportation?

The max tension force for a box is directly related to its stability and safety during handling and transportation. If the max tension force is exceeded, the box could slip or move, causing potential damage to the contents of the box and creating a safety hazard for those handling it. Therefore, it is important to determine and adhere to the max tension force for a box to ensure safe and secure transportation.

5. Can the max tension force for a box change over time?

Yes, the max tension force for a box can change over time due to factors such as wear and tear of the box, changes in the surface it is resting on, and changes in the coefficient of friction. It is important to regularly reassess the max tension force for a box to ensure safe handling and transportation.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
586
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
6K
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top