Finding the minimum force F which must be exerted

AI Thread Summary
The discussion revolves around calculating the minimum force required to prevent a 6 kg block from sliding down an inclined plane on a wedge. The wedge has a mass of 18 kg and a friction coefficient of 0.27, while the surface beneath it is frictionless. Participants emphasize the importance of drawing free body diagrams (FBDs) for both the wedge and the block to accurately analyze the forces at play. Confusion arises regarding the correct representation of normal forces in the FBDs, which are crucial for solving the problem. Overall, the thread highlights the need for clarity in force analysis to find the solution effectively.
rando25
Messages
3
Reaction score
0

Homework Statement


Given: The friction between the block with mass 6 kg and the wedge with mass 18 kg is 0.27 . The surface between the wedge with mass 18 kg and the horizontal plane is smooth (without friction). The acceleration of gravity is 9.8 m/s^ 2. A block is released on the inclined plane (top side of the wedge). What is the minimum force F which must be exerted on the 18 kg block in order that the 6 kg block does not move down the plane?
29zen9h.jpg


Homework Equations


Fnet = ma
and possibly(?): Ff = uFn

The Attempt at a Solution


Having the wedge in the problem is confusing me.
The forces acting on the x-axis of the 18 kg block are F applied and Fg parallel
The forces acting on the x-axis of the 6kg block are F frictional and Fg parallel.
Fnet = ma / Fnet = 0
Fa + Fg(parallel) = 0
I think I am solving for Fa (18kg), I'm not sure.
Fa = Fgsin20
Fa = (18kg)(9.8m/s2)sin20
Fa = 60.33 N
I know the 6kg block plays a role, but i don't know how or where to incorporate it.
I'm really confused, any help is appreciated.
 
Physics news on Phys.org
Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
 
OldEngr63 said:
Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
 
OldEngr63 said:
Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
upload_2014-11-28_18-22-38.png

this is my work so far. I know I am looking of F at this point, but I am still not sure where to go from here? (unless my FBDs are incorrect...)
 
Your FBDs do not agree with each other. For example, the Fn you show on the 6 kg block is upward and to the left; the Fn you show on the wedge is straight up. There should be one Fn that is the opposite of the other. You also have not distinguished the fact that there are two normal forces involved. And it goes on...

I suggest that you try again on the FBDs.
 
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 '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...
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