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

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To hold a 2600kg stone block at rest on a 12-degree incline, the gravitational force acting on the block is calculated to be approximately 24923 N, requiring around 70 workers, each capable of exerting 360 N. For part b, the static friction force of 2560 N must be considered, which reduces the effective force needed to hold the block. The correct approach involves using sine instead of cosine for the calculations related to the incline. Ultimately, the number of workers required will be adjusted based on the frictional force acting against the block's tendency to slide. Accurate calculations are essential for determining the exact number of workers needed.
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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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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