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liq [111]
4 years ago
8

A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

ss m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
1- Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.Express your answer in terms of the variablesM ,m , ms, and appropriate constants.2-A horizontal rope pulls a 10 kg wood sled across frictionless snow. A 6.0 kg wood box rides on the sled. What is the largest tension force for which the box doesn't slip? Assume that Mk= 0.50.
Physics
1 answer:
Nesterboy [21]4 years ago
5 0

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

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Answer:

<h2> 0.041kg</h2>

Explanation:

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Step two:

Required

mass m

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6 0
3 years ago
A jet of water squirts out horizontally from a hole on the side of the tank as shown below. 0.579 m 1.45 m h If the hole has a d
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Answer:

The height of the water above the hole in the tank is 58 mm

Explanation:

In order to solve this problem we need to draw a sketch of the dimensions that include the input variables of the problem.

Where:

x = 0.579[m]

y = 1.45 [m]

Using the following kinematic equation we can find the time that takes the water to hit the ground, and then with this time, we can find the velocity of the water in the x-component.

y = y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\

It is necessary to clarify the value of each of the respective variables below

y = - 1.45 [m] "It is negative because this point is below the water outlet"

yo = 0

vo = 0 "The velocity is zero because the component of the speed on the Y-axis does not exist"

therefore:

-1.45=0.5*(-9.81)*t^{2} \\t = \sqrt{\frac{1.45}{0.5*9.81} } \\t = 0.543[s]

The next step is to determine the velocity in component x, knowing the time.

v=\frac{x}{t} \\v=\frac{0.579}{.543} \\v = 1.06[m/s]

Now using torricelli's law we can find the elevation.

v=\sqrt{2*g*h} \\h=\frac{v^{2} }{2*g} \\h=\frac{1.06^{2} }{2*9.81} \\h= 0.057[m] = 57.95[mm]

3 0
3 years ago
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