Solving Force on a Block Problem

In summary, the block lifts off the ground when the force is large enough. The acceleration is constant, and the velocity is increasing up to the right.
  • #1
joemama69
399
0

Homework Statement



A block is pulled along a rough surface by the force as shown in attachment

1)draw freebody diagram

2)write an expression for the normal force

3)write an expression for coefficient of kinetic friction

4)sketch graph of v vs. t & x vs. t

5)if the force is large enough, the block will lift of ground. find expression for the greatest acceleration that the block can have and still maintain contact with surface

Homework Equations





The Attempt at a Solution



1)FBD ill just explain

Left = uN
Right = FcosQ
Down = mg
Up = N

2) N = mg - FsinQ

3)F(friction) = uN = u(mg - FsinQ)

u = F/(mg-FsinQ)

4) ill explain

v vs t ... I am not sure, is the acceleration constant

x vs t ... the is a slanted line coming out of the origin to the right & up

5) FsinQ - mg = ma

a = (FsinQ-mg)/m
 

Attachments

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  • #2
Part 1) is incorrect; you are neglecting something. I suggest identifying each force that acts on the block, and then, for each force, identify its x and y components. Then, determine what are the x and y components of the acceleration.

Parts 2) and 3) seem to be correct, somehow, inspite of part 1).

Part 4): is the mass constant? is the force constant? is x vs. t a straight line?

You need to think about part 5) some more. You might get a better idea after you fix part 1).
 
  • #3
1) should the up be N - FsinQ

4) mass is constant. force is constant. so the acceleration is constant.

then the velocity is the increasing line up to the right, and displacement is the right side of a parabola
 
  • #4
joemama69 said:
1) should the up be N - FsinQ
Close, but no cigar. Which way does N point: up or down? Which way does F point: up or down?

joemama69 said:
4) mass is constant. force is constant. so the acceleration is constant.
Brilliant induction, Watson!

joemama69 said:
then the velocity is the increasing line up to the right, and displacement is the right side of a parabola
Superb!
 
  • #5
both go up so... N + FsinQ
 
  • #6
joemama69 said:
both go up so... N + FsinQ
Now you're talkin'.
 
  • #7
should it be cos

you use sin for the x, and cos for y correct
 
  • #8
You are using sin for y (up/down) and cos for x (left/right). There is no rule that dictates this (although it is typical); it is the choice that you have made. Is your diagram exactly as given, or should there be an angle defined on the diagram. That might make a difference.

What about part 5)?
 

1. What is the first step in solving a force on a block problem?

The first step in solving a force on a block problem is to draw a free-body diagram of the block. This diagram should show all the forces acting on the block, including any applied forces, weight, and any other external forces.

2. How do I determine the net force acting on the block?

The net force acting on the block can be found by summing up all the forces in the free-body diagram. This includes both the forces in the horizontal and vertical direction. The net force will determine the acceleration of the block.

3. What is the difference between static and kinetic friction?

Static friction is the force that prevents an object from moving when a force is applied to it, while kinetic friction is the force that acts on an object when it is already in motion. Static friction is typically greater than kinetic friction.

4. Can I ignore the weight of the block when solving a force problem?

No, the weight of the block must always be considered when solving a force problem. It is an external force acting on the block and can affect the net force and acceleration of the block.

5. How do I solve for the acceleration of the block?

The acceleration of the block can be found by using Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. By rearranging this equation, you can solve for acceleration by dividing the net force by the mass of the block.

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