What is the solution to the garden cart torque problem?

[akan]

Homework Statement

A garden cart loaded with firewood is being pushed horizontally when it encounters a step 8.0 cm high, as shown in the figure. The mass of the cart and its load is 56 kg, and the cart is balanced so that its center of mass is directly over the axle. The wheel diameter is 60 cm.

Homework Equations

F = mg cos (theta)

The Attempt at a Solution

The angle theta upon which the force is going to act when the cart goes up the slope is:

(R -h) / (sqrt((R - h)² + (h^2))

Multiplied by mg:

mg * (R -h) / (sqrt((R - h)² + (h^2)) =
56 * 9.8 * (.3-.08)/sqrt((.3-.08)^2 + .08^2) = 515.75686308

Rounded to two significant figures, this is 520.

Mastering physics says the answer is 510. How come?

 

[tiny-tim]

A garden cart loaded with firewood is being pushed horizontally when it encounters a step 8.0 cm high, as shown in the figure. The mass of the cart and its load is 56 cm, and the cart is balanced so that its center of mass is directly over the axle. The wheel diameter is 60 cm.

Hi akan!

erm … what is the question asking you to find?

 

[akan]

Hi akan!

erm … what is the question asking you to find?

Oh! That would be an important piece of information, wouldn't it?

Q: What is the minimum horizontal force that will get the cart up the step?

 

[tiny-tim]

may the force be with you …

Oh! That would be an important piece of information, wouldn't it?

ah! the force is strong in this one! ​

(R -h) / (sqrt((R - h)² + (h^2))

What triangle did you get that from?

Hint: draw the forces on the axle, take moments about the step, and choose the correct triangle! (or, at least, one that actually exists )

Mastering physics says the answer is 510. How come?

I get 510 also.

 

[akan]

Ok, thanks, I think I got this one. I set the pivot point at the edge of the cliff, and the arm from that to the center...