Solving a Frictionless Incline Problem: Calculating Normal Force

In summary, a horizontal force of 77 N is pushing a 12 kg block up a frictionless incline at a 60 degree angle with the horizontal. The normal force exerted by the incline on the block can be found by considering all forces with a perpendicular component to the surface, including the weight and the push force. The normal force is equal to the weight's cosine component, which is calculated to be 58.86 N.
  • #1
ur5pointos2sl
96
0
The problem states:

A horizontal force of 77 N pushes a 12 kg block up a frictionless incline that makes an angle of 60 degrees with the horizontal.

What is the normal force that the incline exerts on the block?

After drawing the FBD I can see there are only 2 forces acting in the Y direction which are Fn and W.

so Fn = mg cos theta = 12 kg ( 9.81 m/s^2) * cos (60 deg) = 58.86 N.

This is a homework problem and when attempting to submit the problem on webassign it keeps telling me it is incorrect. I am not sure where I am going wrong or if there is an error with the problem.

Thanks.
 
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  • #2
The 58.86 N looks good for the normal component of the weight.
Now find and add the normal component of the push.
 
  • #3
ur5pointos2sl said:
After drawing the FBD I can see there are only 2 forces acting in the Y direction which are Fn and W.
How are you defining the Y direction? Perpendicular to the surface?

There are three forces acting on the block. What are their directions?
so Fn = mg cos theta = 12 kg ( 9.81 m/s^2) * cos (60 deg) = 58.86 N.
To find Fn you must consider all forces that have a component perpendicular to the surface: All three forces must be considered.
 
  • #4
Doc Al said:
How are you defining the Y direction? Perpendicular to the surface?

There are three forces acting on the block. What are their directions?

To find Fn you must consider all forces that have a component perpendicular to the surface: All three forces must be considered.

Yes the positive Y direction is perpendicular to the surface of the incline. The positive X direction is parallel to the force pushing or pulling the block.

The three forces acting on the block are the Normal force Fn(positive y, No x component), The pull or push force F(positive x, No y component), and also the Weight of the block W(has 2 components, one in the x, one in the y).

Therefore, I would think the only force contributing would be the Weights cos 60 component.

Delphi51 said:
The 58.86 N looks good for the normal component of the weight.
Now find and add the normal component of the push.

When you say the normal component of the push. If I am taking the x direction to be parallel to the push then it would have no normal force. Am I misinterpreting the problem?
 
  • #5
ur5pointos2sl said:
When you say the normal component of the push. If I am taking the x direction to be parallel to the push then it would have no normal force. Am I misinterpreting the problem?
The push force is horizontal, not parallel to the surface. So it will have an x and a y component.
 

1. What is a frictionless incline problem?

A frictionless incline problem is a type of physics problem that involves calculating the forces acting on an object on an inclined surface with no friction. This type of problem is often used to study the effects of gravity and normal force on an object's motion.

2. How do you calculate normal force in a frictionless incline problem?

To calculate the normal force in a frictionless incline problem, you can use the formula FN = mg cosθ, where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.

3. What is the role of the normal force in a frictionless incline problem?

The normal force is the force that acts perpendicular to the surface of an object in contact with another surface. In a frictionless incline problem, the normal force is responsible for balancing out the force of gravity and keeping the object in equilibrium on the incline.

4. How does the angle of the incline affect the normal force in a frictionless incline problem?

The angle of the incline has a direct effect on the normal force in a frictionless incline problem. As the angle increases, the normal force decreases, and vice versa. This is because the force of gravity becomes more perpendicular to the incline as the angle increases, resulting in a smaller normal force needed to balance it out.

5. Are there any real-world applications of frictionless incline problems?

Yes, frictionless incline problems can be applied to real-world scenarios, such as calculating the normal force on an object on a ramp or determining the optimal angle for a ramp to push an object up. These types of problems are commonly used in engineering and physics to understand the effects of forces on objects in motion.

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