Horizontal force with a crate at a 27 degree angle

In summary, to find the required horizontal force F, you can use the angle of the ramp (27°) and the fact that the crate is being pushed at a constant speed to set up equations for the vertical and horizontal components of the applied force. Then, using substitution, you can solve for the unknown variables and find the required horizontal force.
  • #1
bearhug
79
0
A 100 kg crate is pushed at constant speed up the frictionless 27° ramp shown in the figure. What horizontal force F is required?
What force is exerted by the ramp on the crate?

I can't paste the figure up so I hope this question describes the scenario well enough.

What's giving me problems is how am I suppose to find F when I don't even know n? I figured I use the angle to help solve for others. I know Fg=mg which is (100)(9.8) right? Would that give me one side of the triangle?
 
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  • #2
The problem says at a constant speed, therefore a=0. Also, there is no friction slowing you down. F=ma, a=0, so the net force is zero. You need to find the vertical and horizontal components of the applied force. So F(applied)*cos(27)=F(applied,horizontal), and F(applied)*sin(27)=F(applied, Vertical).

That should get you off to a good start, remember you don't know F(applied), so you'll have to solve for two variables (substitution in otherwords).
 

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  • #3
I'm still confused. First of all did you post a figure cause it's not showing up. I understand that I need to use substitution to solve for one and then solve for the other but I still feel like I don't have enough information. This is what I have down:

Xcomponent
∑Fx= max
mgsin27 = ma
a=gsin27

Ycomponent
∑Fy= may
N-mgcos27= 0
N= mgcos27

Is this at all right?
 
  • #4
In the y-sum you forgot to add F, so you have mgcos27 + Fsin27 = N.
 
  • #5
bearhug said:
A 100 kg crate is pushed at constant speed up the frictionless 27° ramp shown in the figure. What horizontal force F is required?
Is the applied force horizontal? or parallel to the ramp?
 

1. What is horizontal force with a crate at a 27 degree angle?

Horizontal force with a crate at a 27 degree angle refers to the amount of force exerted on an object that is being pushed or pulled horizontally while at an angle of 27 degrees from the ground.

2. How is horizontal force with a crate at a 27 degree angle calculated?

The horizontal force with a crate at a 27 degree angle can be calculated using the formula F = mgcosθ, where F is the force, m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of 27 degrees.

3. What factors affect the horizontal force with a crate at a 27 degree angle?

The factors that affect the horizontal force with a crate at a 27 degree angle include the mass of the crate, the angle at which it is being pushed or pulled, and the coefficient of friction between the crate and the surface it is being pushed or pulled on.

4. How does the angle of 27 degrees impact the horizontal force with a crate?

The angle of 27 degrees determines the direction in which the force is being applied and affects the amount of force needed to move the crate. As the angle increases, the horizontal force required to move the crate also increases.

5. What are the practical applications of understanding horizontal force with a crate at a 27 degree angle?

Understanding horizontal force with a crate at a 27 degree angle is important in many real-world scenarios, such as moving heavy objects, designing structures, and calculating the force needed for machines to function efficiently.

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