Which Formula Calculates Gravity's Force on a Block on a Ramp?

  • Thread starter Thread starter J_o_S
  • Start date Start date
  • Tags Tags
    Block Ramp
AI Thread Summary
The discussion centers on determining the correct formula for calculating the force of gravity acting on a block on a ramp. Participants clarify that the question specifically asks for the force of gravity, which leads to the conclusion that the appropriate formula is F = mg. There is some confusion regarding whether to use F = mgsin(theta), but it is established that this formula applies to the component of gravitational force acting parallel to the ramp. Ultimately, the consensus is that for the total gravitational force, F = mg is the correct answer. The discussion confirms the importance of accurately interpreting the question to apply the right formula.
J_o_S
Messages
18
Reaction score
0
What is the force of gravity acting on the block?


are they asking for

a) F = mg

or

b) F = mgsin(theta)

thanks
 
Physics news on Phys.org
J_o_S said:
What is the force of gravity acting on the block?


are they asking for

a) F = mg

or

b) F = mgsin(theta)

thanks

Is that exactly how the question is phrased:What is the force of gravity acting on the block? If so, then I'd say the answer is I'd say a) F=mg.
 
yes that is exactly how it is phrased.

and that is the answer I went with, just wanted to make sure
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top