Solve Forces Problem: Friction & Motion | 15kg Steel Block on Table

• Cathartics
In summary, the conversation is about a problem with a 15 kg block of steel on a horizontal table and the calculation of the force needed to put the block on the verge of moving. The coefficient of static friction and different angles of applied force are also discussed. The person asking for help ultimately figures out the solution three hours later.
Cathartics
forces problem(urgent urgent help needed)

A 15 kg block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is 0.50.
(a) What is the magnitude of the horizontal force that will put the block on the verge of moving?

(b) What is the magnitude of a force acting upward 60° from the horizontal that will put the block on the verge of moving?

(c) If the force acts down at 60° from the horizontal, how large can its magnitude be without causing the block to move?

Come on Carthics, you've been here long enough now to know that you need to show your attempts...

THanks but no thanks hootenanny! I got the answer. And i am a beliver of live and let live and let others prosper.. Thats why i will post the solution here.. for other people who might need it.

a) Force of frcition,
f = μ.N = μmg = 0.5x15x9.8 = 73.5 N
(b) The vertical component of the force, upward reduces effective thrust down.
Normal reaction, N = mg - F sin θ
Force of limiting friction,
fs = μN = 0.5x15x9.8 - 0.5xF sin 60
= 73.5 - 0.433F
This must be equal to the horizontal component of the applied force,
F cos 60 = 0.5 F
Therefore,
0.5 F = 73.5 - 0.433 F
0.933 F = 73.5
F = 73.5/0.933 = 78.8 N

(c) The vertical component of the force, downward increases effective thrust down.
Normal reaction, N = mg + F sin θ
Force of limiting friction,
f's = μN' = 0.5x15x9.8 + 0.5xF sin 60
= 73.5 + 0.433F
This must be equal to the horizontal component of the applied force,
F cos 60 = 0.5 F
Therefore,
0.5 F = 73.5 + 0.433 F
0.067 F = 73.5
F = 73.5/0.067 = 1097 N

Cathartics said:
THanks but no thanks hootenanny! I got the answer.
Well done. However, you managed to work out the solution three hours after posting, so it begs that question, why did you really ask for help? Did you actually need help, or could you just not be bothered to attempt it yourself?

1. What is a force problem?

A force problem is a physics problem that involves the calculation of forces acting on an object, as well as the resulting acceleration and motion of the object. It typically involves identifying all the forces acting on an object and using Newton's laws of motion to solve for unknown quantities.

2. How do I approach solving a force problem?

The first step is to draw a free body diagram, which shows all the forces acting on the object. Then, use Newton's second law (F = ma) to calculate the net force on the object. Finally, use the appropriate equations to solve for the unknown quantities.

3. What are some common types of forces in a force problem?

Some common types of forces include gravity, normal force, friction, tension, and applied forces. In some cases, you may also need to consider non-contact forces such as electric or magnetic forces.

4. How do I handle forces in different directions?

When dealing with forces in different directions, it is important to break them down into their components. This involves using trigonometry to determine the horizontal and vertical components of the force, which can then be used in calculations.

5. What are some real-world applications of force problems?

Force problems are used in various fields, such as engineering, aerospace, and sports. They can be used to design structures, calculate the trajectory of projectiles, and analyze the performance of athletes. They are also essential in understanding the mechanics of everyday objects and activities, such as driving a car or throwing a ball.

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