Equilibrium and Torque and minimum mass

[tambourine]

Homework Statement

To make larger pictures, an enlarger can be moved to the edge of a table so that the image is projected onto the floor, as shown.

mass of bass = 2.0 kg
mass of support = 2.0 kg
mass of enlarger = 6.0 kg

a) What is the minimum mass that needs to be added on the base at point A to keep the enlarger from tipping off the table?

b) With this mass in place, what will be the reaction (normal) force at B?

Homework Equations

T is Torque, F is Force

T = 0
F = 0

The Attempt at a Solution

base (b)
support (s)
enlarger (e)

if A is the pivot and counterclockwise is positive:

-Tb - Ts - Te = 0

did i set the question up correctly? i don't know if A is the pivot point, or if I've got the negatives right.

this torque business is difficult.

 

[PhanthomJay]

You can choose any point about which to sum torques. You chose point A, which is fine. But your plusses and minuses are off. And you don't show the distances from the forces to the chosen pivot point. And you don't show all the forces. There's the weight of the base, the weight of the support, the weight of the enlarger, and the Normal force that acts up at point B. Each of these forces produce a torque about point A. By summing these torques and setting them equal to 0, you can solve for N, the Normal force, then solve for the force required at A by noting that the sum of all forces in the vertical direction must be 0. Alternatively, you could solve for A by summing torques about B. Note also that the weight forces act thru their center of mass. Please show some work so that we can be of further assistance.

 

[tambourine]

thanks jay.

here's what I've got:

i changed the pivot to the bottom of the support (right above B)--(easier to calculate angles and r)

Fs = mg = 2(9.81) = 19.62
Fe = 6(9.81) = 58.86
Fb = 2(9.81) = 19.62



normal is n

T = rFsinθ where r is the distance from the pivot to where the force acts and θ is the angle between r and F

Tb - Ts - Te + Tn = 0

0.15(19.62) - 0.4(19.62) - 0.8(58.86)sin20 + ___?

if the bottom of the support is the pivot, what would the distance to the normal force be?

 

[PhanthomJay]

thanks jay.

here's what I've got:

i changed the pivot to the bottom of the support (right above B)--(easier to calculate angles and r)

Fs = mg = 2(9.81) = 19.62
Fe = 6(9.81) = 58.86
Fb = 2(9.81) = 19.62



normal is n

T = rFsinθ where r is the distance from the pivot to where the force acts and θ is the angle between r and F

Tb - Ts - Te + Tnwhen choosing B as your pivot, which is a good move, you should note that since you are asked to find the minimum mass required at A to prevent overturning, the Normal force acts up directly at point B, that is, the system is just on the verge of tipping over, such that all the normal force acts at that pivot point. Therefore, the Normal force creates no torque about point B. But the mass at A does create a torque. That's what you are trying to solve.[/color]= 0

0.15(19.62) - 0.4(19.62)you forgot to multiply by sin 20 here [/color] - 0.8(58.86)sin20 + ___? This last term should be the torque produced by the mass at A. Watch your plus and minus signs and solve for the weight at A.[/color]

if the bottom of the support is the pivot, what would the distance to the normal force be?

See my comments above in red. You're getting there...

 

[tambourine]

Tb - Ts - Te + Ta = 0

0.15(19.62) - 0.4(19.62)sin20 - 0.8(58.86)sin20 + 0.3Fa = 0

before i go any further, are these the correct positive/negative signs?

 

[PhanthomJay]

Tb - Ts - Te + Ta = 0

0.15(19.62) - 0.4(19.62)sin20 - 0.8(58.86)sin20 + 0.3Fa = 0

before i go any further, are these the correct positive/negative signs?

Yes, looks real good, good job. Now just solve for the force at A, then you need to find the normal force at B.

 

[tambourine]

solving A:

-15.846 + 0.3Fa = 0
0.3mg = 15.846
9.81m = 52.82
m = 5.4

for the normal force:

F = 0
Fn = Fa + Fb + Fs + Fe
Fn = 52.82 + 19.62 + 19.62 + 58.86
Fn = 150.9

the normal force only has a vertical component, right?

 

[PhanthomJay]

Yes, that looks perfect. Normal forces act perpendicular to the surfaces on which they act, and as such, it only has a vertical component in this problem. Looks like you're ready now to get into the torque business!

 

[tambourine]

great! thanks

i will probably be back though. :shy: