M*g = 45*9.8 = 441 N. = Wt. of block.
Fp = 441*sin A = Force parallel with incline.
Fn = 441*Cos A = Normal force.
a. Fs = u*Fn = 0.7*441*Cos A = 308.7*Cos A = Force of static friction.
Fki = u*Fn = 0.59*441*Cos A = 260.2*Cos A = Force of kinetic friction.
b. Fp-Fs = M*a.
441*sin A-308.7*Cos A = M*0 = 0,
441*sin A/Cos A-308.7 = 0,
441*Tan A = 308.7,
A = 35 Degrees.
c. 441*sin37-260.2*Cos37 = M*a.
45a = 57.6,
a = 1.28 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, respectively.
The acceleration of gravity is 9.8 m/s^s. What is the frictional force acting on the 45 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 37 ◦? Answer in units of m/s^2.
1 answer
1 answer