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2 years ago

13

A uniform metal bar of length 6m and mass 100kg rest with its upper end against a smooth vertical wall and with its lower end on

a rough surface of coefficient of friction 0.32.what is the maximum angle made with the horizontal to which the bar can be enclined without sliding?

1 answer:

sdas [7]2 years ago

7 0

Answer:

Let F1x = force of wall on ladder

F1x = F2x the force of friction (balances horizontal forces)

F2y = W balance weight of ladder

F2x = μ F2y force of friction on ladder

F1x L sin θ = W L / 2 cos θ balance torque about point F2

tan θ = W / (2 F1x) previous equation

Now μ W = μ F2y = F2x = F1x from above

So tan θ = W / (2 μ W) = 1 / (2 μ ) = 1 / .64

tan θ = 1.56

θ = 57.4 deg

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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

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The position of the third charge is $x = -1.220 \ m$

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The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

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$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

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substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

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The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

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