In a given experiment, a force F is applied on an object at an angle ? with the

Question

In a given experiment, a force F is applied on an object at an angle ? with the horizontal, as shown in the figure below,

Both the force and the angle are measured, and the results are quoted as a central value plus/minus an uncertainty:

We would like to evaluate the component in the x direction.

1. Let us first conentrate on the central value. Take F0 = 325 N and ?0 = 57 degrees. Find the ventral value of Fx, in N.

2. Both the angles and the magnitude of the forces have a certain uncertainty: eF = 21 N and e? = 1 degrees. Using the propagation methods, calculate the corresponding propagated uncertainty for Fx, in N.

A second force is now applied to the box:

The central values of the measurements are:

F1 = 450 N   F2 = 320 N
?1 = 31 degrees ?2 = 44 degrees

3. What is the central value of the x-component of the net force on this system, in N?

4. For both forces and for both angles, the uncertainty is eF = 8 N and e? = 0.8 degrees. Find the corresponding uncertainty in the x-component of the net force on this system, in N.

Hints:

* Propagate the uncertainty for the x-component of each force first.

* Then, propagate through the subtraction (and becareful with the sign you should use)

* Keep at least 3 decimals in all the intermediate steps.

Explanation / Answer

Part 1 Fx = F * cos 57 = 177 N

When error is included we have

Fx = (325+ 21) *cos(56) is the maximum value = 193.48

and (325-21) cos58 is the minimum value=161.09N

So Fx = 177 +/-   16

Part 2

450 cos 31 = 385.725N; Max value = 458 cos(31-0.8) = 388.923 and min value = 442 cos(31.8) = 375.652

320 cos44 = 230.188N Max value = 328 cos( 44-0.8) = 239.101 and min value = 312 cos( 44.8) = 221.386

Now F1 -F2

Cental value is 155.537   and Max value is 388.923 - 221.386 [ this is because when negative sign max becomes min and min becomes max)

So Max value = 167.537N

Miv value = 375.652-239.101 = 136.551N

So The result is 155.537 + 12 or - 20

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