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Ivahew [28]
3 years ago
6

A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the

figure above. Let Fg be the gravitational force exerted by Earth on the block, Fn be the normal force exerted by the wall on the block, an Ff be the frictional force exerted by the wall on the brick. Which of the following statements about the magnitude of the forces on the block must be true? Select two answers.
A.) F = Fg / Sin(theta)
B.) Fcos(theta) = Fn
C.) Fsin(theta) = Fg + or - Ff
D.) F = Fg + Fn + or - Ff

Physics
1 answer:
wel3 years ago
3 0

Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

Where:

F_{n} is the Normal force

F is the magnitude of the force exerted on the block

\theta is the angle

Net Force in Y:

F sin\theta \pm F_{f}-F_{g}=0 (2)

Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

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the branch currents are as follows:

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  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

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_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

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