In the previous problem, block m1 weighs 874 N. The coefficient of static fricti

Question

In the previous problem, block m1 weighs 874 N. The coefficient of static friction between the block and the table is 0.26 and the angle theta is 34.6. Find the maximum weight of block m2 for which block m1 will remain at rest.

12.[1pt] In the Figure, a block mi sits on a table. There is static friction between block and table. Block m2 hangs from a knot, as shown. Call the tension in the rope connecting the knot to mi, "T1". Call the tension in the rope connecting the knot to m2, "T2". Call the tension in the third rope (the one tipped up by an angle theta, connecting the knot to the wall), T3". The system is in equilibrium Which of the following statements are true? (If A and E are true, and the others are not, enter TFFFT) m2 A) The net force on the knot is downwards. B) The tension T2 must equal m2 g C) If the table was completely frictionless, the system as shown could still be in equilibrum. D) T3 is equal in magnitude to T2 E) The force of static friction on ml equals Tl

Explanation / Answer

(A) net force is zero. False

(B) T2 = m2 g : True

(C) then T1 = 0 and system will not be in equilibrium/

False

(D) T3 sin(theta) = T2

so T3 and T2 are not equal.

False

(E) on m1, Fnet = T1 - f = 0

=> T1 = f

True.

13. T3 cos34.6 = T1 (in horizontal)

T3 sin34.6 = T2 (in vertical)

tan34.6 = T2 / T1

( T2 = m2 g and T1 = f = u N = u m1 g )

tan34.6 = W2 / u W1

W2 = (874 x 0.26 x tan34.6)

W2 = 156.7 N ....Ans

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