Simple Physics Problem (pushing a crate up a slope)

In summary, the coefficient of static friction between the crate and the incline is 0.260. To prevent the crate from sliding down the incline, a minimum force of 38.1 N must be applied perpendicular to the incline. This is found by breaking down the weight of the crate into components and using the coefficient of static friction to determine the necessary force. However, the conversation also mentions that the components may have been broken down incorrectly, leading to the incorrect answer of 21.59 N. The person suggests drawing a free-body diagram to help solve the problem.
  • #1
smashbrohamme
97
1
The coefficient of static friction between the m = 2.80 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Attached is a picture.

I broke down the mass into 27.44 weight, which is not perpendicular to the slope so I broke down the 27.44 into components of 15.75 facing perpendicular towards the slope, and 22.5 going down the slope.

coefficient of static friction is .260(N) N= 15.7 since it is opposite of the weight.

so coefficient ends up being 4.1. so its Force should equal atleast 4.1+22.5. which is 26.6.

the answer is 38.1...what am I doing wrong here? Please help.
 
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  • #2
it seems I broke down my components backwards, but I am still getting the wrong answer.

N=22.5, so friction is 5.85.

15.74+5.85= 21.59, not 38.1 :(
 
  • #3
Did you draw a free-body diagram?
 

1. How do I calculate the force required to push a crate up a slope?

To calculate the force required to push a crate up a slope, you will need to know the weight of the crate and the angle of the slope. The formula for force is force = mass x acceleration. To find the mass of the crate, you can use the formula mass = weight / gravity. Then, you can use the formula force = mass x acceleration to calculate the force needed to push the crate up the slope.

2. How does the angle of the slope affect the force needed to push a crate?

The steeper the slope, the more force will be required to push a crate up it. This is because the steeper the slope, the more the weight of the crate will act against the force of gravity pulling it down the slope. As the angle of the slope increases, the force required to overcome this weight and push the crate up the slope will also increase.

3. What role does friction play in pushing a crate up a slope?

Friction is the resistance between two surfaces that are in contact with each other. When pushing a crate up a slope, friction between the crate and the slope will act against the force being applied to move the crate. This means that a greater force will be needed to overcome the friction and successfully push the crate up the slope.

4. Can you use the same formula for pushing a crate up a slope regardless of its weight?

No, the formula for calculating the force needed to push a crate up a slope takes into account the weight of the crate. A heavier crate will require more force to overcome its weight and move up the slope. However, the angle of the slope will also affect the force needed, as a steeper slope will require more force regardless of the weight of the crate.

5. How does the force needed to push a crate up a slope change if the slope is not perfectly smooth?

If the slope is not perfectly smooth, there will likely be more friction between the crate and the slope. This means that a greater force will be needed to push the crate up the slope, as more force will be required to overcome the increased friction. Additionally, any bumps or obstacles on the slope may also require additional force to be applied to successfully move the crate up the slope.

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