- #1
PhysicsPhun
- 55
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A painter wishes to know whether or not she can safely stand on a ladder. The ladder has a mass M1 = 12 kg which is uniformly distributed throughout its length L = 7.4 m. The ladder is propped up at an angle theta = 53o. The coefficient of static friction between the ground and the ladder is mus = 0.42, and the wall against which the ladder is resting is frictionless. Calculate the maximum mass of the painter for which the ladder will remain stable when she climbs a distance d = 6.6 m up the ladder. (The painter's mass might be so low that only Lilliputian painters can safely ascend the ladder.)
I think that I've gotten the problem but my answer is wrong. Just curious if anyone could take the time to look over my formulas and work.
M_L = 12 kg
L= 7.4m
Theta = 53 degrees
Static Friction (mus) = .42
d = 6.6m
I drew my force diagrams and such, I have a the Normal force of the wall against the top of the ladder, The normal force of the ground against the bottom of the ladder (straight up), Static friction pointing towards the wall. The weight of the ladder at the center of mass of the ladder. And the weight of the Painter/Man.
For this to be in equilibrium, Net Torque and Net force must equal Zero.
I suppose this problem is more tough on the algebraic side?
Here's some of my work.
Positive torque - counterclockwise
Negative torque - Clockwise
Goal - To find the mass of the maximum mass of the man.
Torques:
(L/2)*M_L*g Cos(theta)
d*M_man*g*Cos(theta)
X force N_w= mus*Normal Force
Y force (M_man + M_ladder)g = Normal Force
-L*mus*(M_man + M_L)g*sin(Theta)
So the formula i made to find the mass of the man was this.
(L/2)*M_L*g Cos(theta) + d*M_man*g*Cos(theta) - L*mus*(M_man + M_L)g*sin(Theta) = 0
After plugging in numbers i got
M_man(38.92 - 291.9019) = - 261.8617529
I got 1.035 kg. This is incorrect.. What am i doing wrong?
I think that I've gotten the problem but my answer is wrong. Just curious if anyone could take the time to look over my formulas and work.
M_L = 12 kg
L= 7.4m
Theta = 53 degrees
Static Friction (mus) = .42
d = 6.6m
I drew my force diagrams and such, I have a the Normal force of the wall against the top of the ladder, The normal force of the ground against the bottom of the ladder (straight up), Static friction pointing towards the wall. The weight of the ladder at the center of mass of the ladder. And the weight of the Painter/Man.
For this to be in equilibrium, Net Torque and Net force must equal Zero.
I suppose this problem is more tough on the algebraic side?
Here's some of my work.
Positive torque - counterclockwise
Negative torque - Clockwise
Goal - To find the mass of the maximum mass of the man.
Torques:
(L/2)*M_L*g Cos(theta)
d*M_man*g*Cos(theta)
X force N_w= mus*Normal Force
Y force (M_man + M_ladder)g = Normal Force
-L*mus*(M_man + M_L)g*sin(Theta)
So the formula i made to find the mass of the man was this.
(L/2)*M_L*g Cos(theta) + d*M_man*g*Cos(theta) - L*mus*(M_man + M_L)g*sin(Theta) = 0
After plugging in numbers i got
M_man(38.92 - 291.9019) = - 261.8617529
I got 1.035 kg. This is incorrect.. What am i doing wrong?