** o masses are connected over an inclined plane as shown. N Y. X mi m₂g. = mig TNT MASS 1 ΣFix = FfS N₁ = f 111₂ mag N In the above figure, the inclined surface if frictionless. For parts a) - e), there is no motion. a) In the above figure, (NOT HERE) label the forces on mass 1 (4 forces) and mass 2 (3 forces). b) What are the (x,y) components of mag in the rotated coordinate system? Get numbers for these. N. NY X c) Write down equations for the (x,y) components of the forces on each mass. Remember for this part masses are stationary, so what is a? magy The upper surface has friction. The inclined surface is frictionless. m₁ = 4 kg m₂ ==> 5 kg 0= 40° ΣFiy= ΣF₁ = d) Solve both y equations for the normal forces on each mass. N₂ MASS 2 ΣF2x = N.

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o masses are connected over an inclined plane as shown.
N
↑
X m₁
m₂g. =
↓
mig.
MASS 1
ΣFix =
NT
N₁ =
f
In the above figure, the inclined surface if frictionless. For parts a) - e), there is no motion.
a) In the above figure, (NOT HERE) label forces on
1 (4 forces) and mass 2 (3 forces).
b) What are the (x,y) components of mag in the rotated coordinate system? Get numbers for these.
1T=_
m₂ NY
m₂gy =
c) Write down equations for the (x,y) components of the forces on each mass. Remember for this part
masses are stationary, so what is ax?
X
N.
The upper surface has friction.
The inclined surface is frictionless.
m₁ = 4 kg
m₂ = 5 kg
0 = 40°
ΣF₁y =
ΣFay =
d) Solve both y equations for the normal forces on each mass.
N₂ =
MASS 2
ΣF2x =
N.
e) Solve the F2x. equation for the tension in the string. (Get a number for this).
f) Solve the Fix equation to find the static friction force, then the coefficient of static friction.
Transcribed Image Text:o masses are connected over an inclined plane as shown. N ↑ X m₁ m₂g. = ↓ mig. MASS 1 ΣFix = NT N₁ = f In the above figure, the inclined surface if frictionless. For parts a) - e), there is no motion. a) In the above figure, (NOT HERE) label forces on 1 (4 forces) and mass 2 (3 forces). b) What are the (x,y) components of mag in the rotated coordinate system? Get numbers for these. 1T=_ m₂ NY m₂gy = c) Write down equations for the (x,y) components of the forces on each mass. Remember for this part masses are stationary, so what is ax? X N. The upper surface has friction. The inclined surface is frictionless. m₁ = 4 kg m₂ = 5 kg 0 = 40° ΣF₁y = ΣFay = d) Solve both y equations for the normal forces on each mass. N₂ = MASS 2 ΣF2x = N. e) Solve the F2x. equation for the tension in the string. (Get a number for this). f) Solve the Fix equation to find the static friction force, then the coefficient of static friction.
Telecom
1) Two masses are connected over an inclined plane as shown.
X m
MASS 1
ΣFix =
ΣFly =
Now, say the coefficient of static friction, p, is not as large as the value in part f). So the masses will
be sliding, which means kinetic friction must be used. Use k = .3 for this part of the problem.
From part c), the y equations will not change, but there will be new x equations. Now, ax 0.
g) Determine the new x equations for the moving masses. Just copy the y equations.
h) What is the force of kinetic friction?
fk =
m₂
alx =
N
T =
k = 3 top, horizontal surface only
The inclined surface is frictionless.
j) Use your value for ax to solve for the tension, T.
N
m₁ = 4 kg
m₂ = 5 kg
0= 40°
MASS 2
i) What is aix in terms of a2x? Add the 2 x equations to eliminate T, and solve for the aix
m/s²
ΣΕ =
ΣF2y = 1
Transcribed Image Text:Telecom 1) Two masses are connected over an inclined plane as shown. X m MASS 1 ΣFix = ΣFly = Now, say the coefficient of static friction, p, is not as large as the value in part f). So the masses will be sliding, which means kinetic friction must be used. Use k = .3 for this part of the problem. From part c), the y equations will not change, but there will be new x equations. Now, ax 0. g) Determine the new x equations for the moving masses. Just copy the y equations. h) What is the force of kinetic friction? fk = m₂ alx = N T = k = 3 top, horizontal surface only The inclined surface is frictionless. j) Use your value for ax to solve for the tension, T. N m₁ = 4 kg m₂ = 5 kg 0= 40° MASS 2 i) What is aix in terms of a2x? Add the 2 x equations to eliminate T, and solve for the aix m/s² ΣΕ = ΣF2y = 1
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