Physics help!!, please write out each steps Thank you. If we determine that the
ID: 1485289 • Letter: P
Question
Physics help!!, please write out each steps Thank you.
If we determine that the object will not move so that friction is in operation, then this static frictional force must be strong enough to exactly counteract the horizontal component of the applied force. Only thus we can find how strong the static frictional force actually is If two instead determine that the object will move, then kinetic friction comes into play opposing the motion. We can calculate how strong the kinetic frictional force is by using the coefficient of kinetic friction and the normal force N which we found for that particular situation. A 205 N crate starts out at rest on level ground. The coefficient of static friction between the crate and the ground is 0.50. the coefficient of kinetic (sliding) friction, between the crate and the ground is 0.30. Find the magnitude and the direction of the frictional force acting on the create and the crate's acceleration in each of the following situations.Explanation / Answer
A) if F = 100 N
then Net force Fnet = F-f
f is the frictional force = mu_s*m*g = 0.5*980 = 490 N
So the bodt does not move soaccelaration is zero
if F = 490 N
Fnet = F-f = 490-490 = m*a
accelaration a = 0 m/s^2
if F = 500 N
Fnet = 500-(0.3*980)= 206 N = m*a = (980/9.8)*a
accelaration a = 206*9.8/980 = 2.06 m/s^2
B)
if F = 800 N
then Normal force is N = W+ (800*sin(60)) = 980+(800*sin(60)) = 1672.8 N
frictional force is f = mu_s*N = 0.5*1672.8 = 836.4 N
net force Fnet = m*a =
applied force is 800*cos(60) = 400N
400 < 836.4 then accelaration a = 0 m/s^2
if theta = 30 degrees
applied force is F = 800*cos(30) = 800*0.866 = 692.8 N
frictional force = mu_s*N = 0.5*(980+(800*sin(30))) = 690 N
Since 692.8 > 690 N
so blcok moves
accelaration a = Fnet / m = [(692.8)-(0.3*(980+(800*sin(30))))] / (980/9.8) = 278.8 /100 = 2.78 m/s^2
C) normal force is N = W-F*sin(60) = 980-(500*sin(60)) = 547 N
frictional force F = mu_s*N = 0.5*547 = 273.5 N
applied force is 500*cos(60) = 250 N
since 273.5 > 250 N
the block does not slide
So accelaratiuon a = 0 m/s^2
for theta = 30 degrees
N = W-F*sin(30) = 980-(500*0.5) = 730 N
static frictional force =- mu_s*N = 0.5*730 = 365 N
applied force = 500*cos(30) = 433 N
kinetic frictional force = 0.3*730 = 219 N
then net force Fnet = m*a = 433-219
accelaration a = 214/100 = 2.14 m/s^2