Holding a block of mass in equilibrium on a slope

In summary, you are trying to solve for the weight of a block of mass on an incline, and you are confused because the force of gravity pulling it downward is not the same as the force that is pushing it up the incline.
  • #1
adam19325
2
0

Homework Statement


A block of mass is held in equilibrium on an incline of angle 30 degrees by the horizontal force 500N. Determine the blocks's weight, ignore friction.

Homework Equations


F = MA
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)

The Attempt at a Solution


WlLTfvI.jpg

I'm just wondering if this solution is correct. I'm worried because I think the angle is not the same for the horizontal force as the force of gravity pulling it downward.
 
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  • #2
It is the forces along the incline that need to balance. You need to get correct expressions for the component of the applied force and for the gravitational force component along the incline. There is no requirement for the forces perpendicular to the incline, because the incline will supply the necessary force to balance any applied and/or gravitational forces in this direction.
 
  • #3
adam19325 said:

Homework Statement


A block of mass is held in equilibrium on an incline of angle 30 degrees by the horizontal force 500N. Determine the blocks's weight, ignore friction.

Homework Equations


F = MA
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)

The Attempt at a Solution


WlLTfvI.jpg

I'm just wondering if this solution is correct. I'm worried because I think the angle is not the same for the horizontal force as the force of gravity pulling it downward.
express correctly the angles (between the forces).
For example what is the angle between horizontal push and the inclined plane?
Secondly the block is not sliding so net force along the incline must vanish-
you may draw a free body diagram?[/QUOTE]
 
  • #4
So would it be 500*cos(30) = mg * sin(30)
 
  • #5
adam19325 said:
So would it be 500*cos(30) = mg * sin(30)

It seems to be correct - actually the angle made by F with perpendicular to the incline is (90 -30) degrees.
 
  • #6
adam19325 said:
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)
I don't know what situation those equations apply to, but it's not this one. There is no value in remembering equations separately from the context in which they apply.
 

Related to Holding a block of mass in equilibrium on a slope

1. How does the angle of the slope affect the equilibrium of a block of mass?

The angle of the slope directly affects the equilibrium of a block of mass. As the angle of the slope increases, the force of gravity acting on the block also increases, making it more difficult to maintain equilibrium. At a certain angle, known as the critical angle, the block will start to slide down the slope due to the force of gravity being greater than the force of friction keeping it in place.

2. What is the role of friction in keeping the block of mass in equilibrium on a slope?

Friction plays a crucial role in keeping the block of mass in equilibrium on a slope. Friction acts in the opposite direction of motion, preventing the block from sliding down the slope. Without friction, the block would easily slide down the slope due to the force of gravity.

3. How does the mass of the block affect its equilibrium on a slope?

The mass of the block also plays a role in its equilibrium on a slope. A heavier block will experience a greater force of gravity, making it more difficult to maintain equilibrium. However, a heavier block may also have more surface area in contact with the slope, increasing the friction and making it easier to maintain equilibrium.

4. What other factors can affect the equilibrium of a block of mass on a slope?

Aside from the angle of the slope, friction, and mass of the block, other factors that can affect equilibrium include the surface material of the slope and the shape/size of the block. A rougher surface material will provide more friction, while a smoother surface will provide less. The shape and size of the block can also affect the distribution of weight and the contact area with the slope, impacting the equilibrium.

5. What is the formula for calculating the force of gravity on a block of mass on a slope?

The formula for calculating the force of gravity on a block of mass on a slope is Fg = mgsinθ, where Fg is the force of gravity, m is the mass of the block, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the slope. This formula takes into account the angle of the slope, as well as the mass of the block.

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