# Work done pushing a refrigerator up an Incline

• rorte721
In summary, the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer when using the ramp. The equation 1800 = F(1.5)cos(0) produces the correct answer when lifting straight up.

#### rorte721

Homework Statement
Two workers load identical refrigerators into identical trucks by different methods. One worker has the refrigerator lifted upward onto the back of the truck, which is 1.5 meters above the ground. The other worker uses a ramp to slide the refrigerator onto the back of the truck. The ramp is 5.0 meters long, and raises the refrigerator 1.5 above the ground. The amount of work done by both workers is the same: 1800 J. What are the magnitudes of the forces each worker must exert to load the refrigerators?
Relevant Equations
W=Fdcos(theta)
My teacher told me that answers are F=1200 N for lifting straight up and F= 360 N for using the ramp. I can get the force for lifting straight up by using 1800=F(1.5)cos(0) but I do not understand why for using the ramp the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer.

rorte721 said:
I can get the force for lifting straight up by using 1800=F(1.5)cos(0) but I do not understand why for using the ramp the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer.
In both cases, work is F*D, where D is the displacement in the direction of the force. What is that displacement when sliding it up the ramp?

The displacement parallel to ramp is 5 meters and I understand how 1800=F(5)cos(0) produces the correct answer but I do not understand why I can not use the applied force and the vertical displacement along with the angle between them (72.5) to get the same answer.

rorte721 said:
I do not understand why for using the ramp the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer.
Let us examine the equation you have proposed: 1800 = F(1.5)cos(72.5)

This is an attempt to apply the formula: Work = Force * Distance. The work on the left hand side of the equal sign is correct. So far, so good. The F on the right hand side of the equal sign is our unknown force. Also good.

Edit: I think @Doc Al more correctly interpreted your intent here. You were trying to apply Work = Force * Displacement * cos(angle-between-force-and-displacement). But the relevant displacement is along the ramp, not vertically upward.

Now you appear to want to multiply the Force by the distance moved. Right idea. Wrong implmentation. Here we run into trouble and a lot of unnecessary work.

You want to know the length of the ramp. You know how far it rises. If you knew its angle, you could figure out its length. What is its angle?

You can find the angle by taking the vertical rise divided by the ramp length. That's the cosine of the angle from the vertical. So you divide 1.5 by 5 to get 0.30 and take the inverse cosine of that to get 72.5 degrees. [I assume that is the approach you took].

So now you are back to trying to compute the length of the ramp based on the vertical rise and the angle from the vertical.

You should have applied Rise / Length = cos ( theta ) and solved for length. But if you tried this, you solved it incorrectly.

However, this is all just wasted effort. The length of the ramp is given. [In effect, you have been laboring to take the cosine of an inverse cosine].

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rorte721 said:
The displacement parallel to ramp is 5 meters and I understand how 1800=F(5)cos(0) produces the correct answer but I do not understand why I can not use the applied force and the vertical displacement along with the angle between them (72.5) to get the same answer.
The applied force is parallel to the ramp. The displacement is also parallel to the ramp. (Angle = 0)

You can use a bit of trig to find the parallel displacement in terms of the vertical displacement, but you won't be multiplying by cos(72.5).

jbriggs444
How would that look?

rorte721 said:
How would that look?
Assuming you are responding to my post (but please read @jbriggs444's post): You have a right triangle and you are trying to find the hypotenuse given one of the shorter sides and an angle. So set up your trig using that triangle.

(Of course, you are given the hypotenuse, but you can still do the exercise for fun. )

jbriggs444

## 1. What is the definition of work done pushing a refrigerator up an incline?

The work done pushing a refrigerator up an incline refers to the amount of force exerted over a distance to move the refrigerator against the force of gravity and up the incline.

## 2. How does the incline affect the work done pushing a refrigerator?

The incline affects the work done pushing a refrigerator by changing the direction of the force required to move the refrigerator. It increases the distance over which the force must be exerted, thereby increasing the amount of work done.

## 3. Is it easier or harder to push a refrigerator up an incline compared to on a flat surface?

It is easier to push a refrigerator up an incline compared to on a flat surface because the incline reduces the amount of force needed to move the refrigerator. This is due to the fact that the incline reduces the force of gravity acting on the refrigerator.

## 4. What factors affect the amount of work done pushing a refrigerator up an incline?

The factors that affect the amount of work done pushing a refrigerator up an incline include the angle of the incline, the weight of the refrigerator, and the force applied to move the refrigerator.

## 5. How is the work done pushing a refrigerator up an incline calculated?

The work done pushing a refrigerator up an incline can be calculated by multiplying the force applied to move the refrigerator by the distance it is moved up the incline. This can be represented by the equation W = Fd, where W is the work done, F is the force applied, and d is the distance moved up the incline.