The box contains these masses: (100g, 50g, 10g) = a total of 160 grams, and on t

Question

The box contains these masses: (100g, 50g, 10g) = a total of 160 grams, and on the hanger is hanging these masses: (200g, 20g, 20g) = a total of 240 grams. That's 400 grams of total load. Describe how you could transfer 40 grams from the box to the hanger, while keeping the total mass constant, using only the masses which are already present in the experiment. (Hint: you may need to move masses in both directions.) What minimum acceleration should masses have in this experiment? (in m/s^2)

Theory: A free body diagram is presented below for a box being accelerated across a wooden board by a hanging mass. accelerates this way Figure 2: Free body diagram for a box being accelerated by a hanging mass. m,b = mass of the box, m,-mass of the hanging mass, N= Normal force, T= Tension in the string, f= frictional force Applying Newton's Second Law to each object. Solve for the acceleration of the system in terms of m mh, g, and MT Box Hanging mass = m,a (la, lb) F = mba X axis Y axis (2c) (2d) Substituting equation (2d) into (2a) gives T-ukmhg = mba Adding equation (2c) to equation 3 eliminates T and gives Where mT = m, t mb Solving for the acceleration of the system gives

Explanation / Answer

Only possible of doing this is: move 50 and 10 gram from box to the hanger ; also move 20 g from hanger to box

After shifting of 40 g from box to hanger, mass of the box = 120 g

mass of the hanger = 280 g.

therefore after applying Newton's II Law , we get

mh*g - T = mh*a -------------- (1)

T - u*mb*g = mb * a --------------(2)

mh*g - u*mb*g = (mh + mb)*a

On substitution, we get minimum acceleration (a) = 4 m/s2

Note: Here g = 10 m/s2 and for 'a' to be minimum, u = 1.0 (approximately)

Answer may vary if g = 9.81 m/s2 is taken. Hence amin = 3.924 m/s2

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