Friction problem of stationary block

In summary: That's half the weight of the block. So the mass cancels out.In summary, the problem given involves an initially stationary block of unknown mass on a floor. A force of magnitude 0.500mg is then applied at an upward angle of 20°. The magnitude of the acceleration of the block across the floor is to be determined for two different scenarios: (a) when the coefficients of static and kinetic friction are 0.620 and 0.530, respectively, and (b) when the coefficients are 0.430 and 0.340, respectively. The mass of the block is not given, so it will cancel out during calculations. The first step in solving the problem is to draw a diagram and
  • #1
oscarkool2
8
0

Homework Statement



Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a)μs = 0.620 and μk = 0.530 and (b)μs = 0.430 and μk = 0.340?



Homework Equations



none


The Attempt at a Solution



Problem seems unsolvable to me. No mass given, no force, etc. Have no idea what to do.
 
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  • #2
There is a good chance that at some point during your calculations, the mass will cancel out leaving you with just g somewhere. Either that or they just want your answer in an algebraic form.
 
  • #3
no I need a final answer. I'm sure it will cancel out since it is impossible to solve for but I don't know where to even began for that. Where do I even begin?
 
  • #4
oscarkool2 said:

Homework Statement



Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a)μs = 0.620 and μk = 0.530 and (b)μs = 0.430 and μk = 0.340?



Homework Equations



none


The Attempt at a Solution



Problem seems unsolvable to me. No mass given, no force, etc. Have no idea what to do.

Step Number ONE: Do not try to solve the problem BEFORE you understand the problem!


Don't even think about what to do until you draw yourself a good picture with the forces labeled. Put in a set of coordinate axes. Then identify the x and y components of the forces. Note that in the y direction you have gravity acting down, but there is also a component of the applied force acting up. Note that there is a frictional force acting horizontally in one direction and the x component of the applied force acting in the other (I'm imagining the x-axis being horizontal).

Now ask yourself a couple questions: (1) Is the vertical component of the applied force larger than mg? If yes then the block comes off the ground. If no then it doesn't. (2) Is the horizontal component of the applied force larger than the static frictional force? If yes then the block moves horizontally. If no then the block doesn't move.

If the block moves then you can use

[tex] \Sigma F_x = ma_x [/tex]
and
[tex] \Sigma F_y = ma_y [/tex]

as appropriate.

(By the way, I assume that in your other post you were referring to the textbook by Halliday and Resnick. Thousands of students have learned physics from that text. My own copy dates back to 1966. Oh, and thousands of students have cussed at it too :mad: :smile:)
 
  • #5
Well I don't understand how to solve for the mass. any idea? once i have the mass i can solve for the x and y components and do whatever else i need to do.
 
  • #6
Ahh I think I'm on the right track. mass does end up canceling since I am solving for acceleration. so i assume mass isn't calculated when solving acceleration?

actually I am confused again. I've got a static and a kinetic friction. so how do i determine which friction is used for the x and y component for acceleration to calculate its magnitude?

edit: giving up on the homework for tonight, I'm tired of looking at this stupid problem for 4 hours and getting no where. thanks for the reply but it didn;t help me :(
 
Last edited:
  • #7
oscarkool2 said:
Ahh I think I'm on the right track. mass does end up canceling since I am solving for acceleration. so i assume mass isn't calculated when solving acceleration?

actually I am confused again. I've got a static and a kinetic friction. so how do i determine which friction is used for the x and y component for acceleration to calculate its magnitude?

edit: giving up on the homework for tonight, I'm tired of looking at this stupid problem for 4 hours and getting no where. thanks for the reply but it didn;t help me :(

Since I'm into handing out advice today, working 4 hours on a problem (even if it's not a continuous 4 hours) is a waste of time. One ends up going around in circles. It's better to put it aside and even sleep on it. You may not know it but your brain does a lot of processing while you're sleeping.

Okay, about static and kinetic friction. Here static means "not moving" and kinetic means "moving". Go back to my first response. I said to check to see if the horizontal force was greater than the static frictional force. You compute the static frictional force using the coefficient of static friction. Here's a general rule: If the static frictional force is larger than the component of the applied force acting in the same direction, the object ain't going to move. If, on the other hand the component of the applied force is greater than the static frictional force, the object will move AND then you have to use the coefficient of kinetic friction to compute the frictional force opposing the motion.

Now for your problem there isn't going to be any y component to the acceleration. Why? because the vertical component of the applied force is not large enough to lift the block. The block is going to slide horizontally, period.

One last thing. Your first clue that the mass would cancel out of the equations was that the applied force was given as .500mg.
 

Related to Friction problem of stationary block

1. What is friction?

Friction is a force that resists motion between two surfaces that are in contact with each other. It is caused by the roughness of the surfaces and the interlocking of their microscopic irregularities.

2. How does friction affect a stationary block?

Friction can keep a stationary block from moving if the force applied to the block is not strong enough to overcome the force of friction. This is known as static friction.

3. What factors affect the amount of friction on a stationary block?

The amount of friction on a stationary block is affected by the roughness of the surfaces, the weight of the block, and the normal force (the force perpendicular to the surfaces).

4. Is there a way to reduce friction on a stationary block?

Yes, friction can be reduced by using lubricants, such as oil or grease, on the surfaces in contact. This creates a layer between the surfaces, reducing the amount of direct contact and therefore reducing the friction.

5. How is friction beneficial?

Friction is beneficial in many ways. It allows us to walk without slipping, drive cars, and hold objects. In some cases, friction can also be used to slow down or stop objects in motion, such as brakes on a car or the grip on a bicycle tire.

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