A block of mass m/ = 4.0 kg is put on top of a block of mass m2 = 5.0 kg. To cause the top
block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at
least 12 N must be applied to the top block. The blocks assembly is now placed on a
horizontal, frictionless table, as shown in FIGURE 1. Find the magnitudes of
a.the maximum horizontal force that can be applied to the lower block so that the blocks
will move together
b.the resulting acceleration of the blocks.
3.
- The maximum horizontal force that can be applied to the lower block is 5 N.
- The resulting acceleration of the blocks are 21,1 m/s².
Penjelasan dengan langkah-langkah:
Diketahui:
Massa 1 = 4 kg
Berat 1 = 40 N
Massa 2 = 5 kg
Berat 2 = 50 N
F1 = 12 N
Ditanya:
- Gaya pada blok paling bawah (F2)?
- Percepatan (a)?
Jawab:
∑Fy1 = 0
N1 – W cos θ = 0
12 – 120 cos θ = 0
12 = 120 c0s θ
cos θ = 0,1
∑Fx1 = m1 x a
120 – T = 4a….(1)
∑Fy2 = 0
N2 – w cos θ = 0
N2 = w cos θ
N2 = 50 (0,1) = 5 N
∑Fx2 = m2 x a
50 sin θ + T = 5a
50 x 1 + T = 5a
50 + T = 5a……….(2)
120 – T = 4a
50 + T = 5a
___________ +
190 = 9a
a = 21,1 m/s²
from that calculation, we know that:
- The maximum horizontal force that can be applied to the lower block is 5 N.
- The resulting acceleration of the blocks are 21,1 m/s².
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