If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?

Engineering

1 answer:

sleet_krkn [62]2 years ago

7 0

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".