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mjcox2000 wrote:

Unome wrote:

I haven't seen a problem this difficult in nearly a year, and I've never seen a problem like this before, so I'll probably get it wrong, but... (Click to Show Hidden Text)

FL cos 60 degrees is the torque on the bent arm, which should be equal to the torque 6 J/rad on the other side (I don't know how gravity affect this) and somehow that 32.0 N force on the fulcrum forms another equation... this isn't really getting me anywhere. Nice problem though!

blindmewithscience wrote:

I'm not really sure, but I think that this is the right way to go off of Unome's part (Click to Show Hidden Text)

Maybe the 32N force is acting as a "normal force" for the system as a whole? So the mass weighs 32-20=12N? I've also not seen something like this, and I'm not sure if this is correct at all...

You're both right about what you've gotten so far, and together, you're pretty close to the solution.

If the whole lever system isn't moving (i.e. is still) then both the net linear forces and net torques acting on it must be zero. 
Now, starting with the forces, we see that there is a force downward of 20N and a force upward of 32N. With only one other place a force can act (at F), the force at F must be downward 12N to make the net force on the system 0N.
Next, looking at the net torques, we can now disregard the 32N force upward as it is at the fulcrum (F*D would be 0 newton-meters as the distance is 0); this leaves solely the force of F (12N) acting on a lever arm of Lcos60 to balance the torque on the right of 30*20=600. Solving this equation, we get L=100cm.

JonB wrote:If the whole lever system isn't moving (i.e. is still) then both the net linear forces and net torques acting on it must be zero.

Unome wrote:

JonB wrote:If the whole lever system isn't moving (i.e. is still) then both the net linear forces and net torques acting on it must be zero.

Oh, that's new to me (The forces part).

chinesesushi wrote:

Unome wrote:

JonB wrote:If the whole lever system isn't moving (i.e. is still) then both the net linear forces and net torques acting on it must be zero.

Oh, that's new to me (The forces part).

LOL really? I'm just saying, that's like the first rule of static equilibrium calculations, that net forces and net torques both = 0

JonB wrote:My turn for a question:
What is the angle of the inclined plane in the diagram below such that the weight of 50kg is lifted with a constant velocity? (All the numbers on the pulleys are radii, and μ is the coefficient of kinetic friction)

compound problem 1.png

I'll assume that the pulleys are massless.... :roll: 
Since the 50kg weight is being lifted with a constant velocity, the net force on it must equal 0. The net downward force, 490N, must be equally distributed to the two strings holding the weight up. Thus, the tension in each string pulling up must be 245N. 

Accounting for the forces on the 200kg block, there is the component of the gravitational force pushing to the lower left, there is the tension from the string pulling to the upper right, and there is the force of friction in the direction opposite of motion, in this case the upper right. Again, the forces directed to the upper right must cancel out with the forces directed to the lower left since the 200kg block is moving at constant velocity. 

Forces to the upper right = Tension + Friction = 245N + (mu)(Normal Force) = 245N + (0.2)(200*9.8*cos(theta)) = 245N + 392cos(theta)
Forces to the lower left = Component of Force of Gravity = 200*9.8*sin(theta) = 1960sin(theta)
245 + 392cos(theta) = 1960sin(theta) 
Using a graphing calculator, theta is approximately 0.320277 radians or 18.3505 degrees.

lchs wrote:

JonB wrote:My turn for a question:
What is the angle of the inclined plane in the diagram below such that the weight of 50kg is lifted with a constant velocity? (All the numbers on the pulleys are radii, and μ is the coefficient of kinetic friction)

compound problem 1.png

Solution...maybe (Click to Show Hidden Text)

I'll assume that the pulleys are massless.... :roll: 
Since the 50kg weight is being lifted with a constant velocity, the net force on it must equal 0. The net downward force, 490N, must be equally distributed to the two strings holding the weight up. Thus, the tension in each string pulling up must be 245N. 

Accounting for the forces on the 200kg block, there is the component of the gravitational force pushing to the lower left, there is the tension from the string pulling to the upper right, and there is the force of friction in the direction opposite of motion, in this case the upper right. Again, the forces directed to the upper right must cancel out with the forces directed to the lower left since the 200kg block is moving at constant velocity. 

Forces to the upper right = Tension + Friction = 245N + (mu)(Normal Force) = 245N + (0.2)(200*9.8*cos(theta)) = 245N + 392cos(theta)
Forces to the lower left = Component of Force of Gravity = 200*9.8*sin(theta) = 1960sin(theta)
245 + 392cos(theta) = 1960sin(theta) 
Using a graphing calculator, theta is approximately 0.320277 radians or 18.3505 degrees.

245+392cosθ=1960sinθ
5/8+cosθ=5sinθ
5/8=5sinθ-cosθ
sin^2(θ)+cos^2(θ)=1
cosθ=sqrt(1-sin^2(θ))
5/8=5sinθ-sqrt(1-sin^2(θ))
Substitute x for sinθ
sqrt(1-x^2)=5x-5/8
1-x^2=25x^2-25x/4+25/64
26x^2-25x/4-39/64=0
Quadratic formula
x=(25/4±sqrt(1639/16))/52
1639/16>25/4,and sinθ is positive, so eliminate the extraneous solution
x=(25/4+sqrt(1639/16))/52
(25/4+sqrt(1639)/4)/52
(25+sqrt(1639))/208
Re-substitute sinθ for x
sinθ=(25+sqrt(1639))/208
θ=arcsin((25+sqrt(1639))/208)
Evaluate to whatever precision you need - it comes out to about 0.3203 radians.

labchick wrote:

Is it (Click to Show Hidden Text)

the moment of inertia?

mjcox2000 wrote:

lchs wrote:

JonB wrote:My turn for a question:
What is the angle of the inclined plane in the diagram below such that the weight of 50kg is lifted with a constant velocity? (All the numbers on the pulleys are radii, and μ is the coefficient of kinetic friction)

compound problem 1.png

Solution...maybe (Click to Show Hidden Text)

I'll assume that the pulleys are massless.... :roll: 
Since the 50kg weight is being lifted with a constant velocity, the net force on it must equal 0. The net downward force, 490N, must be equally distributed to the two strings holding the weight up. Thus, the tension in each string pulling up must be 245N. 

Accounting for the forces on the 200kg block, there is the component of the gravitational force pushing to the lower left, there is the tension from the string pulling to the upper right, and there is the force of friction in the direction opposite of motion, in this case the upper right. Again, the forces directed to the upper right must cancel out with the forces directed to the lower left since the 200kg block is moving at constant velocity. 

Forces to the upper right = Tension + Friction = 245N + (mu)(Normal Force) = 245N + (0.2)(200*9.8*cos(theta)) = 245N + 392cos(theta)
Forces to the lower left = Component of Force of Gravity = 200*9.8*sin(theta) = 1960sin(theta)
245 + 392cos(theta) = 1960sin(theta) 
Using a graphing calculator, theta is approximately 0.320277 radians or 18.3505 degrees.

Agreed - here's a more rigorous/precise version of the last step (Click to Show Hidden Text)

245+392cosθ=1960sinθ
5/8+cosθ=5sinθ
5/8=5sinθ-cosθ
sin^2(θ)+cos^2(θ)=1
cosθ=sqrt(1-sin^2(θ))
5/8=5sinθ-sqrt(1-sin^2(θ))
Substitute x for sinθ
sqrt(1-x^2)=5x-5/8
1-x^2=25x^2-25x/4+25/64
26x^2-25x/4-39/64=0
Quadratic formula
x=(25/4±sqrt(1639/16))/52
1639/16>25/4,and sinθ is positive, so eliminate the extraneous solution
x=(25/4+sqrt(1639/16))/52
(25/4+sqrt(1639)/4)/52
(25+sqrt(1639))/208
Re-substitute sinθ for x
sinθ=(25+sqrt(1639))/208
θ=arcsin((25+sqrt(1639))/208)
Evaluate to whatever precision you need - it comes out to about 0.3203 radians.