Physics, 21.06.2019 13:30 10040813

Rubber on floor tile has a coefficient of kinetic friction of 1.0 and a coefficient of static friction of 2.0. how much force would a 3.5 lb. robot need to produce in order to start moving?

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The Friction Force $F_{r}$ is the force that exists between two surfaces in contact, which opposes to the relative movement between both surfaces, or the force that opposes the start of the slide. Hence, there are two types:

-The force of kinetic friction: acts when the body is moving with respect to the surface on which it is moving. Note the coefficient of kinetic friction is $\mu_{k}$

-The force of static friction: acts when the body is at rest with respect to the surface on which it is moving. Note the coefficient of static friction is $\mu_{s}$

In this context, we have the Free Body Diagram (FBD) of the situation, as shown in the figure attached. Here we have the robot with a mass on a rubber surface.

In this case there is no vertical motion, this means that the weight is countered by the Normal Force [N], which is equal in magnitude to the weight, but in opposite direction.

In the horizontal plane of the FBD, is shown that the friction force $F_{r}$ that is opposed to the force .

Having this clear, let’s begin:

Firstly, we have to convert the lb to kg, knowing the following:

1lb=453.59237g=0.4535kg (1)

This means:

$m=3.5lb\frac{0.4535kg}{1lb}=1.5875kg$

m=1.5875kg >>>This is the value of the mass in kilograms

On the other hand, we know the weight is:

W=mg

Where is the acceleration due gravity with a value of $9.8\frac{m}{s^2}$

$W=(1.5875kg)(9.8\frac{m}{s^2})$

W=15.55N >>>>This is the weight of the robot

Remembering the weight is countered by the Normal Force, which is equal in magnitude to the weight, but in opposite direction:

W=N=15.55N (2)

Now, the Friction Force is:

$F_{r}=N\mu$ (3)

If we want to move the robot, the Force must be greater than the Friction Force $F_{r}$ :

$FF_{r}$ (4)

For $\mu_{k}$ :

$F_{r}=\mu_{k}N$

$F_{r}=(1)(1.55N)$

$F_{r}=1.55N$ (5)

For $\mu_{s}$ :

$F_{r}=\mu_{s}N$

$F_{r}=(2)(1.55N)$

$F_{r}=31.116 N$ (5)

This means the force excerted on the robot must be greater than 31.116N

F31.116N

Rubber on floor tile has a coefficient of kinetic friction of 1.0 and a coefficient of static fricti

Answer from: loveagirl111puppy

As we know that given mass will be

m = 3.5 lb

1 lb = 0.454 kg[/tex]

now we will have

m = 1.6 kg

now we know that in order to move the mass from its stationary position we need to apply external force which must be more that static friction force

so we need to apply a force which is more than static friction force

F = F_s

$F = \mu_s mg$

As we know that

$\mu_s = 2.0$

now from above equation

F = 2.0(1.6)(9.8)N

F = 31.36 N

so the applied force will be 31.36 N

Answer from: Quest

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

Answer from: Quest

Pe = 1/2k(x)^2 and .05n/m = .05x10^-9m = x

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