simple physics problem: frictional forces....?

Question

simple physics problem: frictional forces....?

Answer 1

to help you i am going to take this problem step by step

the first thing is to make sure the box will move. for the box to move the force applied has to be greater than the force of static friction. if it is not the box will not move. we must first find the force of static fricion. to do that we use the equation us=fs/fn to solve for the fs or force of static, we multiply the coefiiciant of static by the normal force, or wieght of the box. if we do this we get the fs to be 38.81N of force. since the applied force is greater than we know that the box will move. now we hav to find out how fast the box will be accelerating. to do this we have to find the net force acting on the box. the net force is the force applied minus the force of kinetic friction.

so

Fr=Fa-Fk

to find fk we use the equation uk=fk/fn solve for fk by multiplying the coefficient of kinetic friction by the force applied. do this and you should get 27.06N of resistent force.

now we must find the sum of all forces by subtracting the force of friciton from the applied force.

Fr=45.2-27.06

Fr=18.136N

newtons second law tells us that F=ma or force equals mass times accelration. we also know that weight equals mass times gravity. so if you divide the normal force by gravity (9.81m/s^2) it will give you mass. if you do that the box should weigh 5.75kg

since F=ma divide the resultant force by the mass to give you the acceleration.

a=F/m

a=18.136/5.75

a=3.154m/s^2

and that is your final answer.

Answer 2

Since the block is kept horizontally,its normal force is equal to its weight, i.e 56.5N. The minimum force that would be required to move this object is f(s).N,i.e 38.81N, which is less than applied force. In this case kinetic friction would come to play. This force is equal to f(k).N,i.e 27.06N. Therefore net force is 18.14N(45.2-27.06). Mass of the body is 5.76kg(56.5/g). Therefore acceleration is F/m i.e 3.15 m/s2

Answer 3

I keep seeing your questions, so I'll answer one.





..............W = 56.5 N

................↓

45.2N →▓▓▓←Ff

¯¯¯¯¯¯¯¯¯↑¯¯¯¯¯¯

...............N



In the diagram, you'll see that the force of 45.2 N is only opposed by the force of Friction (Ff)



Ff =µN <=== µ here is the static coefficient of friction



Where µ = 0.687 for static friction

..........N = W <===from ΣFv = 0 (no motion along this direction)

Ff = (0.687)(56.5)

...= 38.82 N



Now, the Force 45.2N is greater than Ff of 38.82N,

Therefore the block will move!!!



There is an unbalanced force

So, ( F = ma) will apply





.............W.......===>F

..............↓......==>a

45.2N→▓▓▓←Ff (kinetic)

¯¯¯¯¯¯¯¯↑¯¯¯¯¯¯¯¯¯¯¯

.................N





From Newton's 2nd Law



F = ma

...............▌Where:

...............▌F = 45.2N - Ff

...............▌Ff (kinetic) = µkN

...............▌µk =0.479

...............▌N =56.5newtons



(45.2 - Ff) = ma

(45.2 - µkN) = ( W/g)a

[45.2 - 0.479(56.5)] = [(56.5) / (9.8)]a

[ 45.2 - 27.06] = [5.77]a



a = (18.14) / (5.77) = 3.14 m/sec²