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Georgia [21]
2 years ago
15

The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulle

y
and pulls with a force of magnitude 91.7 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 59.2 N, T2 =
33.4 N, and T3 = 9.65 N. What are the masses of (a) disk A. (b) disk B, (c) disk C, and (d) disk D?
11
79
Physics
1 answer:
garik1379 [7]2 years ago
7 0

Answer:

A) m = F/a = 91.7/9.81 = 9.35 kg

B) m = F/a = 59.2/9.81 = 6.03 kg

C) m = F/a = 33.4/9.81 = 3.40 kg

D) m = F/a = 9.65/9.81 = 0.984 kg

Explanation:

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Read 2 more answers
Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2
Svetllana [295]

Answer:

<u>I had to search the Figure on Google to solve this question.</u>

a) The magnitude of the force F₃ is:

F_{3} = 87.47 N

And the direction of F₃:

\alpha = 79.04 ^{\circ}  (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N  

Explanation:

<u>I had to search the Figure on Google to solve this question.</u>

a) We can find the force of the third person as follows:

\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0

\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0

So, in x-direction we have:

\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0

F_{3x} = -85.87 N

In y-direction we have:

\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0

F_{3y} = -16.63 N

The magnitude of the force F₃ is:

F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}

\alpha = 79.04 ^{\circ}

<em>b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a. </em>      

<u>To find the weight of the cart​ when it accelerates at 200 m/s², we need to consider: F₃ = 0</u>.  

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

\Sigma F_{x} = F_{1x} + F_{2x} = ma

45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2}

m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg

Now, the weight of the cart​ is:

P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N

I hope it helps you!                                                                                    

3 0
2 years ago
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