# How Does Angle of Force Application Affect Acceleration on a Frictional Surface?

• chahud42
In summary, the problem involves finding the acceleration of a block of mass 2.65 kg on a horizontal floor, given an applied force of 81.0 N at an angle of 50° to the vertical, with coefficients of friction μs = 0.895 and μk = 0.665. After breaking the force into components and applying Newton's second law and trigonometry, the normal force and force of friction can be calculated. However, after re-evaluating the calculations, it is determined that the applied force does not overcome the force of static friction, resulting in an acceleration of 0.
chahud42

## Homework Statement

A force is applied to an initially stationary block of mass 2.65 kg that sits on a horizontal floor as shown. The 81.0 N force is applied at θ = 50° angle to the vertical. The coefficients of friction between the floor and the block are μs = 0.895 and μk = 0.665. What is the acceleration of the block?

## Homework Equations

Newton's second law and trig

## The Attempt at a Solution

First I broke the force into components, Fx=81sin50, Fy=81cos50. There are also the force of gravity, normal force, and friction acting on the block. So for sum of forces equations i have
, and
. So
. Knowing normal force, you can substitute that in the equation for x-axis.
. Next I wanted to figure out whether the acceleration is 0 because it doesn't overcome friction or not. So I compared the force of friction and applied force. Force of static friction I worked out to be 46.6N, where the applied force is 81sin50, which is 62N. So, we use kinetic friction...

Where did I go wrong?

How did you arrive at 46.6 Newtons for static force? What value is the normal force? Look back at your formula for Fn, did you forget something?

scottdave said:
How did you arrive at 46.6 Newtons for static force? What value is the normal force? Look back at your formula for Fn, did you forget something?
To find the static force I multiplied .895 by the expression for Fn. I took that as the force of friction to overcome, which the applied force of 62N overcomes, and accelerates the system so that kinetic friction must use. To find normal force, I applied Newtons second law to the y-axis. ∑Fy=0 (acceleration is equal to 0). ∑Fy=Fg+Fappliedy-Fn=0
Sooooo Fn=Fg+Fapplied=mg+81cos50

Am I looking at something wrong?

chahud42 said:
To find the static force I multiplied .895 by the expression for Fn
Sure, but as scottdave hinted you forgot part of your equation when calculating the normal force.

haruspex said:
Sure, but as scottdave hinted you forgot part of your equation when calculating the normal force.
I genuinely have no clue. The only forces in the negative y direction are the force of gravity and the y component of the applied force. So the normal force should be equal to the sum of those.

chahud42 said:
To find the static force I multiplied .895 by the expression for Fn. I took that as the force of friction to overcome, which the applied force of 62N overcomes, and accelerates the system so that kinetic friction must use. To find normal force, I applied Newtons second law to the y-axis. ∑Fy=0 (acceleration is equal to 0). ∑Fy=Fg+Fappliedy-Fn=0
Sooooo Fn=Fg+Fapplied=mg+81cos50

Am I looking at something wrong?
What is the numerical value of the normal force?

chahud42 said:
I genuinely have no clue. The only forces in the negative y direction are the force of gravity and the y component of the applied force. So the normal force should be equal to the sum of those.
You initially wrote, correctly
chahud42 said:
And again post #3:
chahud42 said:
Sooooo Fn=Fg+Fapplied=mg+81cos50
But when you calculated ##\mu F_n## you appear to have forgotten the mg term. If you still cannot get the right answer, please post all working in complete detail.

haruspex said:
You initially wrote, correctly

And again post #3:

But when you calculated ##\mu F_n## you appear to have forgotten the mg term. If you still cannot get the right answer, please post all working in complete detail.
I re-crunched the numbers and got 69.84N for the force of static friction and 62.05 for the applied force, so it seems it doesn't overcome it at all, making the acceleration zero. I can't believe I had it right the whole time. I must have just had it written down without the mg term in my notebook. Thanks for the help.

## What is the concept of pushing a box at an angle?

The concept of pushing a box at an angle involves applying a force to an object at an angle rather than directly in line with the object's motion.

## How does pushing a box at an angle affect its motion?

Pushing a box at an angle will cause the box to move in both the horizontal and vertical directions, resulting in a curved path instead of a straight line.

## What factors affect the force required to push a box at an angle?

The force required to push a box at an angle is affected by the mass of the box, the angle at which the force is applied, and the coefficient of friction between the box and the surface it is being pushed on.

## What are some real-life applications of pushing a box at an angle?

Pushing a box at an angle is commonly used in moving furniture or other heavy objects around a room or up a flight of stairs. It is also used in sports such as billiards, where players must push the cue ball at an angle to make it hit other balls.

## What are some key principles to keep in mind when pushing a box at an angle?

When pushing a box at an angle, it is important to remember to use enough force to overcome the friction between the box and the surface it is being pushed on. It is also important to push in the correct direction and at the correct angle to achieve the desired motion of the box.

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