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Lady_Fox [76]
2 years ago
10

1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)

Physics
1 answer:
kotegsom [21]2 years ago
7 0

As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.

Then   friction force = 80 Newtons              but in the opposite direction.

Friction force =  Mu  * Normal force exerted by ground  =  Mu * weight of box

So we find Mu.

Mu = coefficient of friction between box and horizontal surface

          = Force of friction / weight  =  80 / 50 * 9.81 = 0.163

When an identical box is placed on top, the force of friction is

      = Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons

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A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly
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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m
Bumek [7]

Answer:

a)  total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is  d1=0.5*L

By Steiner theorem

for the rod we get J_1'=J_1 + m_1\times d_1^2

J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2

for the sphere we get J_2' = J_2 + m_2\times d_2^2

J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2

And the total moment of inertia for the first case is

J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2

b) F=476 N

The torque for system is given as

M = F\times d\times sin(a)

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

M = F\times L/2

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

a_1 = \frac{M}{J_{t1}}

      = \frac{1161.44}{1359.05}

      = 0.854 rad/sec^2

4 0
3 years ago
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