Number of Workers Needed to Hold 2600kg Stone Block at Rest

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Homework Help Overview

The discussion revolves around determining the number of workers required to hold a 2600kg stone block at rest on a ramp inclined at 12 degrees. The problem involves concepts from physics, particularly forces, friction, and static equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the force required to hold the block at rest using gravitational force components and questions the correctness of their approach. Some participants suggest revisiting the free body diagram and using sine functions instead of cosine for the calculations. Others inquire about the relationship between the forces in part b and the static friction acting on the block.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of sine in calculations, and there is a consensus on the approach to part b, although explicit consensus on the overall solution has not been reached.

Contextual Notes

Participants are navigating the complexities of static friction and the forces acting on an inclined plane, with some expressing uncertainty about the initial calculations and the implications of friction in the scenario.

TyFelmingham
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Suppose there is a 2600kg stone block on a ramp inclinded at 12 degrees to the horizontal.Assume that each worker can exert a pulling force of magnitude at most 360 Newtons.

a) Determine the number of workers it would take to hold the stone block at rest

b)Although lubrication can reduce friction, it cannot eliminate it. Assume that
the magnitude of the static friction exerted by the ramp on the stone block
is 2560 N. Taking into account the friction acting upwards along the ramp
(reducing the tendency of the block to slide down the ramp), how many men
are required to hold the block at rest?

Ok, so for a,

mg* cos 12
=24923 N

Divide it by the amount a worker can pull

= 70

Is that right?

For b, I am stuck...I have no idea where to start
 
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It should be a sinus there...Make the FBD again & pay attention to those triangles.

Daniel.
 
For b), would the force just be the F calculated in question a minus 2560 N?

Cheers
 
Yes that's how you do part B.
I also agree with dex when he says to use sine instead of cosine.
 

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