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Question

x PROBLEMS x PROBLEMS ?GAStone Of Mass 296 , x'( 36282/take Business Bank New Tab Escherichia coli straMparvicella DNA fo DT Question 1 10 pts The figure shows a box of mass m2 on an incline at an angle 9-33.2. It is connected by a cord that runs over a pulley to a box of mass m1 -2.77 kg on a horizontal surface. All surfaces and the pulley are frictionless. The masses of the cord and the pulley are negligible. A horizontal force F is applied to m1, as shown. What is the largest value the magnitude of F could have (in Newtons) without the connecting cord losing tension? D Question 2 10 pts A ball of mass 2.03 kg is attached to a vertical rod by two strings, as shown in the figure. The rod rotates making the ball move in a horizontal circle with a period of 0.969 s. Find the tension in the upper string and the tension in the lower string. Give the difference between the two tensions (in Newtons). (Note: The tension in the strings depends on the period of rotation, but the difference between the two does not.) 60? 50en ch

Explanation / Answer

Question - 1

For mass m2, the force and resulting acceleration equation is -

m2*g*sin? - T = m2*a ------------------------------(i)

And for the mass m1 -

T + F = m1*a ---------------------------(ii)
solve these equations and get

T = m2*(m1*g*sin? - F)/(m1 + m2)

For finding the maximum value of F, take the limiting case,

Means -

F - m1*g*sin? = 0

=> F = m1*g*sin? = 2.77*9.81*sin33.2 = 14.88 N

Question2 -

Radius of the circle, r = sqrt(50^2 - 30^2) = 40 cm = 0.40 m

Period of rotation, T = 0.969 s

So, frequency f = 1 / T = 1 / 0.969 = 1.032 Hz

So, angular velocity w = 2*pi*f = 2*3.141*1.032 = 6.48 rad/s

So, linear velocity v = w*r = 6.48*0.40 = 2.59 m/s

The force which is keeping the object moving in a circle is the centripetal force = m * v^2/r

So -

Centripetal force = (2.03 * 2.59^2) / 0.40 = 34.13 N

Now, the two strings must exert enough force to make a total of 34.13 N horizontally
The horizontal component of the tension of each string = ½ * 34.13 = 17.06 N

The horizontal component of the tension = T * cos ?
The strings are at an angle from horizontal.
The cosine of the angle = (horizontal distance) ÷ (length of string)

=>  cos ? = 0.40 / 0.50 = 0.80

Therefore we have -

17.06 = T*0.80

=> T = 17.06 / 0.80 = 21.33 N

Therefore, tension in each string = 21.33 N.

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