How To Find The Magnitude Of Friction Force

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Magnitude of frictional force

• Thread starter hansel13
• Start date October 25, 2008

Homework Statement

A 2.5 kg cake is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 Northward and a vertical forcefulness P are and then applied to the block. The coefficients of friction for the block and surface are u_s = 0.twoscore and u_k = 0.25. Determine the magnitude of the frictional forcefulness acting on the cake if the magnitude of P is
(a) 8 N
(b) 10 N
(c) 12 N

Homework Equations

f_k = U_kNorth
Due north = mg

The Attempt at a Solution

part (a)
N = 2.5kg*9.8m/s² = 24.five Due north

So, f_g = 0.25*24.five N = 6.125

But the answer to part (a) was half-dozen.0, non 6.125... Information technology feels like I'm quite a bit off track. Could someone give me a little footstep forward on part (a)?

Answers and Replies

You lot are assuming that the block is actually moving when the horizontal and vertical forces are applied.

Remember, the static frictional force increases as the applied forcefulness on your object increases (just in the opposite direction) until it reaches a maximum [tex]f_{smax}=\mu_s Northward[/tex].

In other words, until the applied force reaches a value equal to that of [tex]f_{smax}[/tex] in that location volition be no motion AND the magnitude of the static frictional strength will be equal to the magnitude of the applied force.

For (a) in your example (taking the negative management of an imaginary y-centrality as positive) when calculating the resultant "normal" force [tex]N_{res}[/tex] on your block (y-axis simply) we get

[tex]N_{res} = F_{mg} + P = (2.5kg \times nine.eight chiliad/south^2) \\- \\8N = 16.5N[/tex]

You know the coefficient of static friction is [tex]\mu_s=0.forty[/tex] and therefore

[tex]f_{smax} = 16.5N \times 0.twoscore = 6.6N[/tex]

This is clearly larger than the horizontal force of six.0N so nosotros tin can conclude that there is no horizontal motion when a vertical force of 8N is applied (since [tex]f_{smax}[/tex] opposes the horizontal force until the horizontal force reaches this maximum value).

Therefore the magnitude of the frictional forcefulness at this indicate must all the same equal to that of the horizontal force, i.eastward. frictional force = 6.0N.

Now yous should just apply the same principles to the remaining parts of your problem and you should be ok.

Promise that helps

Adept stuff!

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