# Sliding box question- finding applied force with friction

• jgray
In summary: Since F is the applied horizontal force, what expression yields Fax (its component parallel to the slope)?
jgray

## Homework Statement

A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
when the ramp is frictionless.
when the coefficient of kinetic friction is 0.18.

Ff= uk(Fn)

## The Attempt at a Solution

I think I have the answer to part 1 complete and correct: I tilted the axis and ended up multiplying sin28degrees(90kg)(9.8m/s^2) to give me 414N for the applied force.
Im not sure about the second part though, where I have friction involved. I tried finding my force of friction --> (0.16)(882N) = 141 N, then subtracting it from my initial applied force I found - so 414N - 141 N gives me 273 N. Not sure if this is correct though.. Help please!

jgray said:

## Homework Statement

A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
when the ramp is frictionless.
when the coefficient of kinetic friction is 0.18.

Ff= uk(Fn)

## The Attempt at a Solution

I think I have the answer to part 1 complete and correct: I tilted the axis and ended up multiplying sin28degrees(90kg)(9.8m/s^2) to give me 414N for the applied force.
Keep in mind that the applied force is horizontal. Only a component of that force will be directed along the slope. Draw a free body diagram and see if you can't work out how to decompose the applied force into components along the slope and normal to the slope.

Im not sure about the second part though, where I have friction involved. I tried finding my force of friction --> (0.16)(882N) = 141 N, then subtracting it from my initial applied force I found - so 414N - 141 N gives me 273 N. Not sure if this is correct though.. Help please!

Again, since the applied force is not parallel to the slope, but is horizontal, it's component normal to the slope will contribute to the net normal force that results in friction...

jgray said:

## Homework Statement

A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
when the ramp is frictionless.
when the coefficient of kinetic friction is 0.18.

Ff= uk(Fn)

## The Attempt at a Solution

I think I have the answer to part 1 complete and correct: I tilted the axis and ended up multiplying sin28degrees(90kg)(9.8m/s^2) to give me 414N for the applied force.
No, you are assuming that the applied force is parallel to the incline, when it is given that the applied force is horizontal.
Im not sure about the second part though, where I have friction involved. I tried finding my force of friction --> (0.16)(882N) = 141 N, then subtracting it from my initial applied force I found - so 414N - 141 N gives me 273 N. Not sure if this is correct though.. Help please!
Since the applied force is horizontal, it has a component perpendicular to the incline, which affects the value of the normal force.
But why in any case would the applied force be less than the no friction case? It would be harder to push , wouldn't it?

Okay I see what you guys mean about the force being parallel versus horizontal . So instead, I break the Fapplied into x and y components. Since my Fnet of x is 0 d/t no acceleration, my Fgx should be equal and opposite to Fax.. So 414N I think. Then I used cos28Fax to find the applied force, so 370N?

jgray said:
Okay I see what you guys mean about the force being parallel versus horizontal . So instead, I break the Fapplied into x and y components. Since my Fnet of x is 0 d/t no acceleration, my Fgx should be equal and opposite to Fax.. So 414N I think. Then I used cos28Fax to find the applied force, so 370N?

You can't get a component of a force that's larger than the force itself (414 is larger than 370) So revisit your calculation of the applied force. If F is the applied horizontal force, what expression yields Fax (its component parallel to the slope)?

## 1. What is the "sliding box question" and why is it important?

The sliding box question is a physics problem that involves finding the applied force needed to move a box across a surface with friction. It is important because it helps us understand the relationship between force, friction, and motion, which is essential in many real-world scenarios.

## 2. How do you approach solving the sliding box question?

To solve the sliding box question, you must first draw a free-body diagram and identify all the forces acting on the box. Then, use Newton's Second Law (F=ma) to set up an equation with the forces and solve for the unknown force. Finally, check your answer by plugging it back into the equation and ensuring that it satisfies all the given conditions.

## 3. What is the role of friction in the sliding box question?

Friction is the force that resists the motion of the box across the surface. It is an essential factor in the sliding box question because it determines the amount of force needed to overcome it and move the box. Without friction, the box would move with a constant velocity, and the question would not be as challenging.

## 4. Is there a simplified version of the sliding box question?

Yes, there is a simplified version of the sliding box question, which involves only one type of force acting on the box. In this case, the applied force is equal to the force of friction, and the equation becomes F=μN, where μ is the coefficient of friction and N is the normal force exerted by the surface on the box.

## 5. What are some real-life applications of the sliding box question?

The sliding box question has many real-life applications, such as determining the force needed to push or pull a heavy object, calculating the maximum weight a person can carry, or designing machines that require precise force calculations. It is also essential in understanding the dynamics of vehicles, such as cars and trains, and predicting their motion and stopping distances.

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