a. Wt. = m * g 25kg * 9.8N/kg = 245 N.
F1 = Force parallel to the incline.
F2 = Force perpendicular to the incline.
Q1. F1 = 245*sinA = u*F2
245*sinA = 0.45*245*cosA
Divide both sides by 245*cosA:
sinA/cosA = 0.45
Replace SinA/cosA with tanA:
tanA = 0.45
A = 24.2o = Angle of the incline.
Fs=u*245*cosA=0.45*245*cos24.2=100.6 N.
= Force of static friction.
Q2. The largest angle is the angle at which the force of static friction is
equal to F1. The angle is 24.2o.
Q3. F1 = 245*sin29 = 118.8 N.
F2 = 245*cos29 = 214.3 N.
Fk = u*F2 = 0.38 * 214.3 = 81.4 N. =
Force of kinetic friction.
a=Fn/m = (F1-Fk)/m=(118.8-81.4)/25 = 1.49 m/s^2.
A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.45 and μk = 0.38, respectively.
The acceleration of gravity is 9.8 m/s2 .
What is the frictional force acting on the 25 kg mass?
Answer in units of N
What is the largest angle which the incline can have so that the mass does not slide down the incline?
Answer in units of ◦
What is the acceleration of the block down the incline if the angle of the incline is 29◦ ?
Answer in units of m/s2
4 answers
this is sd afae[a
24.2o what is the sign after 24.2?
degrees probably
1 answer