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MissTica
4 years ago
6

A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction

is 0.8, what is the weight of the box?
Physics
1 answer:
alukav5142 [94]4 years ago
6 0
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
 (565 / 0.8) = 706.25N. weight.
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A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

7 0
3 years ago
A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of
lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight  (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

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W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

 m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W=  ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7  N

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