Free body diagram(its simple but i don't know why i can't understand it)

  • Thread starter Thread starter 1/2"
  • Start date Start date
  • Tags Tags
    Body Free body
AI Thread Summary
The discussion revolves around calculating the maximum force that can be applied to a 5kg block without causing a 4kg block on top to slide off. The maximum static friction for the top block is calculated as fmax=μ mg, resulting in 12 N using a coefficient of friction μ=0.036. Participants emphasize the need for a free body diagram to visualize the forces acting on both blocks, particularly the interaction between them due to friction. Questions arise about the acceleration of the 5kg block when a force is applied and the implications for the upper block without friction. Understanding these dynamics is crucial for accurately solving the problem.
1/2"
Messages
98
Reaction score
0

Homework Statement


If a 4kg block is placed on top of a 5kg block which placed on a friction less plane and a force is applied on the lower 5kg block ,For what maximum value of force will the top block not slide ?
μ=0.036

Homework Equations


The Attempt at a Solution


The maximum static friction that the top block can experience is fmax=μ mg=12 N
But the force acts on the lower block so what will be the force on the upper block that the friction will counter?
It would be nice to get a free body diagram.
 
Physics news on Phys.org
1/2" said:

Homework Statement


If a 4kg block is placed on top of a 5kg block which placed on a friction less plane and a force is applied on the lower 5kg block ,For what maximum value of force will the top block not slide ?
μ=0.036

Homework Equations





The Attempt at a Solution


The maximum static friction that the top block can experience is fmax=μ mg=12 N
But the force acts on the lower block so what will be the force on the upper block that the friction will counter?
It would be nice to get a free body diagram.

Would you agree that the 5 kg block will have to accelerate if a force is applied horizontally?
 
The frictional acts between the blocks: it is a force of interaction that opposes the relative motion of both blocks. Try to draw the free body diagram separately for both blocks.

What force acts on the upper block? What would happen to it without friction?

By the way, how did you get fmax=μ mg=12 N with μ=0.036?

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top