What is the acceleration of the block up the ramp?

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The discussion centers on calculating the acceleration of a large block being dragged up a ramp by a smaller block over a pulley. The large block has a mass of 1400 kg, while the smaller block is 530 kg, with a friction coefficient of 0.13 and a ramp angle of 15 degrees. Participants highlight the forces acting on the blocks, including gravitational forces and friction, and express confusion over negative acceleration results. Clarification is provided that the entire gravitational force of the smaller block should be considered acting parallel to the ramp. Ultimately, the conversation concludes with a better understanding of the forces involved and the correct approach to solving the problem.
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Homework Statement



A large block is being dragged up a ramp by a block falling due to gravity over a pulley. The large block has a mass of 1400kg and the smaller block suspended has a mass of 530kg. The coefficient of kinetic friction between the large block and the ramp is 0.13. The ramp is at an angle of 15degrees
what is the acceleration of the block up the ramp?

Homework Equations


F=ma
Ff=ukN

The Attempt at a Solution


I know there are four forces actnig on the block, and these are:
in the negative x direction there is the mass of the big block * gravity *sin15 + Ff (m*g*sin15 + 0.13*m*g*cos15)
and in the positive x direction there is the smaller block pulling it down via gravity so m*g*sin15 (m in this case is the smaller block)
a=fnet/m where fnet= mlittle*g*sin15 -(mbig*g*sin15 + 0.13*mbig*g*cos15)
The problem is it ends up being negative when I go to do, and I'm sure the block is mean to be accelerating up the ramp.
a=Fnet/m
 
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PTPM93 said:

Homework Statement



A large block is being dragged up a ramp by a block falling due to gravity over a pulley. The large block has a mass of 1400kg and the smaller block suspended has a mass of 530kg. The coefficient of kinetic friction between the large block and the ramp is 0.13. The ramp is at an angle of 15degrees
what is the acceleration of the block up the ramp?

Homework Equations


F=ma
Ff=ukN


The Attempt at a Solution


I know there are four forces actnig on the block, and these are:
in the negative x direction there is the mass of the big block * gravity *sin15 + Ff (m*g*sin15 + 0.13*m*g*cos15)
and in the positive x direction there is the smaller block pulling it down via gravity so m*g*sin15 (m in this case is the smaller block)
a=fnet/m where fnet= mlittle*g*sin15 -(mbig*g*sin15 + 0.13*mbig*g*cos15)
The problem is it ends up being negative when I go to do, and I'm sure the block is mean to be accelerating up the ramp.
a=Fnet/m

It seems like you have a whole lot of forces jumbled up there. Since no loss due to friction was stipulated for the pulley, I think you can consider the whole force due to gravity of the weight is acting parallel to and up along the ramp. Now work on the friction.
 
This is what I've figured out so far but I don't think I'm getting the right answer
5yhgsg.jpg
 
I seem to be getting ~-2ms^-2 as the acceleration and it should be positive??
I don't understand what I'm doing wrong I have
Net F = ma
Fbigblock = m*g*sin theta - (M*g*sin theta + 0.15 *M * g * cos theta)
.'. a = m*g*sin theta - (M*g*sin theta + 0.15 *M * g * cos theta) divided by M
But it doesn't seem right
 
PTPM93 said:
This is what I've figured out so far but I don't think I'm getting the right answer
5yhgsg.jpg

Attached is diagram of what I said before. The smaller weight merely supplies a constant force along the slope of the ramp. You've made the problem harder than it is.
 

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Ah okay, Thanks a lot I understand now
 

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