Wb = m*g = 29kg * 9.8N/kg = 284.2 N. = Wt. of the block.
1. Fp = 284.2*sin A = Force parallel to the incline.
Fn = 284.2*cos A = Normal or Force perpendicular to the incline.
Fs = us*Fn = 0.7*284.2*cos A=199*cos A
Fp-Fs = m*a = m*0 = 0
284.2*sinA - 199*c0sA = 0
284.2sinA = 199cosA
Divide both sides by cosA:
284.2*(sinA/cosA) = 199
Replace sinA/cosA with tanA:
284.2*tanA = 199
tanA = 188/284.2 = 0.70
A = 35o
Fs = 199*cos35 = 163 N. = Force of static friction.
2. A = 35o.
3. Fp = 284.2*sin40 = 182.7 N.
Fn = 284.2*cos40 = 217.7 N.
Fk = uk*Fn = 0.59*217.7 = 128.4 N. = Force of kinetic friction.
a=(Fp-Fk)/m = 182.7-128.4)=1.87 m/s^2
A block is at rest on the incline shown in the
figure. The coefficients of static and kinetic
friction are μs = 0.7 and μk = 0.59, respec-
tively.
The acceleration of gravity is 9.8 m/s2
What is the frictional force acting on the
29 kg mass?
Answer in units of N
(part 2 of 3)
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Answer in units of �
(part 3 of 3)
What is the acceleration of the block down
the incline if the angle of the incline is 40� ?
Answer in units of m/s2
1 answer
1 answer