Two blocks are positioned on surfaces, each inclined at the same angle theta 67.

Question

Two blocks are positioned on surfaces, each inclined at the same angle theta 67.6 degrees with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of the black block is 1.4 kg, and the coefficient of friction for both blocks and inclines is 1.52. Assume gravity is g=9.8 m/s2.

Picture: white block on the left of the incline and black block on the right

A) What must be the mass of the white block if both blocks are to slide to the right at a constant velocity?

B) what must be the mass of the white block if both blocks are to slide to the left at a constant velocity?

C) What must be the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5 m/s2?

D) What must be the mass of the white block if both blocks are to slide to the left at an acceleration of 1.5 m/s2?

E) Now re-do part (D) above, but now assuming no friction at all on either incline. So in the absence of friction, what must be the mass of the white block such that both blocks slide to the left at an acceleration of 1.5m/s2?

Explanation / Answer

N = m g cos67.6

fk = uk N = 1.52 m g cos67.6 = 0.58 m g

(A) block slide to the right.

friction on right block will be upwrd and on left block will be downward.

constant velocity means a = 0

T + f - m g sin67.6 = 0

and T - f' - m' g sin67.6 = 0

f' + m' g sin67.6 = mg sin67.6 - f

0.58 m' g + 0.924 m' g = 0.924m g - 0.58mg

1.504 m' = 0.344 m

m' = (0.344 / 1.504) (1.4 kg) = 0.32 kg ......Ans

(B) T - f - m g sin67.6 = 0

and T + f' - m' g sin67.6 = 0

m' g sin67.6 - f' = mg sin67.6 + f

0.924 m' g - 0.58 m' g = 0.924m g + 0.58mg

1.504 m = 0.344 m'

m' = (1.504 / 0.344) (1.4) = 6.12 kg .........Ans

(c) T + f - m g sin67.6 = m a

T + (0.58 m g) - (0.924m g ) = 1.5 m

T = 4.87 m

and T - f' - m' g sin67.6 = m' a

4.87m - 0.58 m' g - 0.942m' g = 1.5m'

m' = 0.415 kg .......Ans

(D)

T - f - m g sin67.6 = m a

T - 0.58 m g - 0.942 m g = 1.5 m

T = 16.42 m

and T + f' - m' g sin67.6 = m' a

16.42(1.4) + 0.58 m' g - 0.942 m' g = 1.5 m'

m' = 4.55 kg

(E) m' g sin67.6 - T = m' a

T - 1.4g sin67.6 = 1.4a

m'g sin67.6 - 1.4 g sin67.6 = 1.5 ( m' + 1.4)

9.06 m' - 12.68 = 1.5m' + 2.1

m' = 1.96 kg

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