Calculating the Pulling Force for a 130 kg Block with Friction Consideration

Introduction to the Scenario

A 130 kg block is resting on a horizontal table with a coefficient of kinetic friction of 0.3. Our goal is to determine the force required to pull this block at a constant speed when the applied force is at an angle of 25 degrees above the horizontal. This involves understanding and applying concepts of both static and kinetic friction.

Understanding the Forces Involved

To solve this problem, we need to consider the weight of the block, the applied force, the normal force, and the frictional force.

Weight of the Block

The weight (W) of the block can be calculated using the formula:

W m times; g

where m 130 kg and g 9.81 m/s2.

Substituting the values, we get:

W 130 kg times; 9.81 m/s2 1275.3 N

Forces Acting on the Block

Applied Force (F): The force we want to determine, applied at an angle of 25 degrees above the horizontal. Normal Force (N): The force exerted by the table on the block, perpendicular to the surface of the table. Kinetic Frictional Force (fk): The force that opposes the motion of the block.

Resolving the Applied Force into Components

The applied force can be resolved into a horizontal (Fx) and a vertical (Fy) component:

Horizontal Component: Fx F cos(25°) Vertical Component: Fy F sin(25°)

Calculating the Normal Force (N)

The normal force is the force exerted by the surface of the table on the block, which is affected by the vertical component of the applied force:

N W - Fy W - F sin(25°)

Substituting the weight of the block:

N 1275.3 N - F sin(25°)

Calculating the Kinetic Frictional Force (fk)

The kinetic frictional force is given by:

fk μk times; N

Given that the coefficient of kinetic friction (μk) is 0.3:

fk 0.3 times; N 0.3 times; (1275.3 N - F sin(25°))

Setting Up the Equation for Constant Speed

For the block to move at a constant speed, the horizontal component of the applied force must equal the kinetic frictional force:

Fx fk

Substituting the expressions:

F cos(25°) 0.3 times; (1275.3 N - F sin(25°))

Distribute and rearrange the equation:

F cos(25°) 0.3 F sin(25°) 382.59 N

F (cos(25°) 0.3 sin(25°)) 382.59 N

F 382.59 N / (cos(25°) 0.3 sin(25°))

Substitute the trigonometric values:

cos(25°) ≈ 0.9063

sin(25°) ≈ 0.4226

cos(25°) 0.3 sin(25°) ≈ 0.9063 0.3 times; 0.4226 ≈ 0.9063 0.12678 ≈ 1.03308

Now, substitute back to find F:

F ≈ 382.59 N / 1.03308 ≈ 369.3 N

Therefore, the force required to pull the block at a constant speed is approximately 369.3 N.

Conclusion

Understanding the relationship between the applied force, the normal force, and the frictional force is key to determining the pulling force required for the block to move at a constant speed. This example demonstrates how to apply these principles to a real-world scenario.