physics q

[inaccensa]

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If a rubber object is on a rubber surface, and if the object's mass is increase by adding several weights inside it, which of the following is true of the magnitude of te force req to inititate horizontal movement (Fx) and the weight of the object w? There is friction in this problem both kinetic and static a formula, I dont think is needed to solve this problem is (1/u=F/A) u=coefficient of friction

The ratio of Fx to W

-remains constant
-decreases

Thanks

 

[ScoobyDooo]

If a rubber object is on a rubber surface, and if the object's mass is increase by adding several weights inside it, which of the following is true of the magnitude of te force req to inititate horizontal movement (Fx) and the weight of the object w? There is friction in this problem both kinetic and static a formula, I dont think is needed to solve this problem is (1/u=F/A) u=coefficient of friction

The ratio of Fx to W

-remains constant
-decreases

Thanks

I think this is a trick problem as you don't need work to solve the problem.

In order for the system to move, the force applied (Force of initiation -->) should be greater than the static friction (<---).

Force applied > Force static friction

Force static = u (static) x w (mass of all weights) g

 

[inaccensa]

I think this is a trick problem as you don't need work to solve the problem.

In order for the system to move, the force applied (Force of initiation -->) should be greater than the static friction (<---).

Force applied > Force static friction

Force static = u (static) x w (mass of all weights) g

I dont think you understood my q. they are asking about the ratio of force:weight. I answered constant which is incorrect

 

[circulus vitios]

I dont think you understood my q. they are asking about the ratio of force:weight. I answered constant which is incorrect

Are you sure it's decreases?

 

[inaccensa]

Are you sure it's decreases?

yeah that is the correct ans

 

[costales]

At max static equilibrium, applied force Fx = friction = uN = uW, so Fx/W = u. But static u > kinetic u, so to get into the kinetic regime (to get the object moving) you want smaller u, which means smaller Fx/W.

 

[inaccensa]

At max static equilibrium, applied force Fx = friction = uN = uW, so Fx/W = u. But static u > kinetic u, so to get into the kinetic regime (to get the object moving) you want smaller u, which means smaller Fx/W.

When you put it that way🙂

 

[mikexima]

your original post is poorly worded. I cant comprehend it.

 

[circulus vitios]

your original post is poorly worded. I cant comprehend it.

Same.