Find Least Force to Move 10-ton Safe on Truck

[holy_kamote]

find least force?

Homework Statement

a safe weighing 10 tons is to be loaded on a truck 5.0 ft high by means of planks 20 ft long if it requires 350lb to overcome friction on the skids, find the least force nescessary to move the safe.

Homework Equations

The Attempt at a Solution

i don't know where to start pls help me...

 

[holy_kamote]

what is the formula for finding least force help me pls?

 

[Doc Al]

What force parallel to the planks is required to just overcome gravity and friction?

 

[holy_kamote]

sir doc al will i use the cosine law? thanks for the repz..

 

[holy_kamote]

will i use conversion with this prob? ton??

 

[holy_kamote]

i convert ton in kg?

 

[Doc Al]

Keep everything in terms of pounds. 1 ton = 2000 lbs (at least in the US).

You'll need to find the component of the weight parallel to the incline.

 

[holy_kamote]

i convert 10 tons to 20000 lbs? then i get the angle 5/20=0.25cos=14.50degrees? i calculate force of gravity 20000 x 9.8 = 196000.. then the normal force 196000/cos14.50=184757.7...then i get the parallel to the incline 196000/sin14.50=49074.5 is that correct? what's the next step? pls reply thxxxxx...

 

[Doc Al]

i convert 10 tons to 20000 lbs?

Yes, that's the weight.

then i get the angle 5/20=0.25cos=14.50degrees?

OK. The sine of the angle is 5/20 = 0.25; thus that angle must be 14.5 degrees.

i calculate force of gravity 20000 x 9.8 = 196000..

No, you already have the force of gravity. 20000 lbs is a force, not a mass.

then the normal force 196000/cos14.50=184757.7...then i get the parallel to the incline 196000/sin14.50=49074.5 is that correct?

No, not correct. You have the wrong value for the force. But beyond that, you are calculating the components incorrectly. Read this: http://www.physicsclassroom.com/Class/vectors/U3l1e.cfm#trig"

Once you get the correct parallel component of the weight, you need to add the friction force to get the total force needed.

 

[holy_kamote]

ok normal force= 20000sin14.50=5007.6 and theparallel component 20000cos14.50=19363 is this correct?

 

[Doc Al]

ok normal force= 20000sin14.50=5007.6 and theparallel component 20000cos14.50=19363 is this correct?

You have them mixed up. The angle is with respect to the horizontal, so the normal component of the weight would be Wcosθ and the parallel component would be Wsinθ.