problem with static frictional...

PHYS101 Problem with static frictional force

Problem with static frictional force

The figure below shows a block of mass m connected to a block of mass M = 2.00kg, both on45◦ inclined planes where the coefficient of static friction is 0.28. Find the minimum and themaximum values for m for which the system is at rest.

45° 45°

m M

The masses m and M are on the verge to slide up or down for the minimum and the maximummass of m. If we want to determine the minimum mass, obviously the mass M will be on theverge to slide down and the mass of m will be on the verge to slide up. Therefore the staticfrictional force for M will be parallel to the tension of the cord, whereas the the static frictionalforce for m will be antiparallel to the tension of the cord and also antiparallel to the potentialdirection of the motion.Let us now consider the Free Body Diagrams for m and M in the case of the minimum massof m.

From the Free Body Diagram for m weget:

∑ Fy = Fn −mg cos θ = 0

⇒ FN = mg cos θ

∑ Fx = T −mg sin θ − fs = 0

with fs = µsFN we get:

T −mg sin θ − µsmg cos θ = 0 (1)

From the Free Body Diagram for M weget:

∑ Fy = Fn −Mg cos θ = 0

⇒ FN = Mg cos θ

∑ Fx = −T −Mg sin θ − Fs = 0

with Fs = µsFN we get:

−T + Mg sin θ − µsmg cos θ = 0 (2)

c©2014 Department of Physics, Eastern Mediterranean University Page 1 of 2

PHYS101 Problem with static frictional force

(1) +(2) gives−mg(sin θ + µs cos θ) + Mg(sin θ − µs cos θ) = 0

Solving this equation for m gives:

m =sin θ − µs cos θ

sin θ + µs cos θM =

sin 45◦ − 0.28 cos 45◦

sin 45◦ + 0.28 cos 45◦2.00kg = 1.125kg

If we want to determine the maximum mass, obviously the mass M will be on the verge toslide up and the mass of m will be on the verge to slide down. Therefore the static frictionalforce for M will be antiparallel to the tension of the cord, whereas the the static frictional forcefor m will be parallel to the tension of the cord and also parallel to the potential direction ofthe motion.Let us now consider the Free Body Diagrams for m and M in the case of the maximum massof m.

From the Free Body Diagram for m weget:

∑ Fy = Fn −mg cos θ = 0

⇒ FN = mg cos θ

∑ Fx = T −mg sin θ+ fs = 0

with fs = µsFN we get:

T −mg sin θ+µsmg cos θ = 0 (3)

From the Free Body Diagram for M weget:

∑ Fy = Fn −Mg cos θ = 0

⇒ FN = Mg cos θ

∑ Fx = −T −Mg sin θ+Fs = 0

with Fs = µsFN we get:

−T + Mg sin θ+µsmg cos θ = 0 (4)(1) +(4) gives

−mg(sin θ−µs cos θ) + Mg(sin θ+µs cos θ) = 0

Solving this equation for m gives:

m =sin θ+µs cos θ

sin θ−µs cos θM =

sin 45◦+0.28 cos 45◦

sin 45◦−0.28 cos 45◦2.00kg = 3.28kg

c©2014 Department of Physics, Eastern Mediterranean University Page 2 of 2