A block with mass m1 = 9.4 kg is on an incline with an angle = 33° with respect

Question

A block with mass m1 = 9.4 kg is on an incline with an angle = 33° with respect to the horizontal. For the first question there is no friction between the incline and the block

4.The spring is replaced with a massless rope that pulls horizontally to prevent the block from moving. What is the tension in the rope?

5.Now a new block is attached to the first block. The new block is made of a different material and has a coefficient of static friction = 0.97. What minimum mass is needed to keep the system from accelerating?

Explanation / Answer

4) Assuming µ = 0.374,
we still have the downslope acceleration g*sin
but now upslope we have the rope tension component
(T/m)cos
and a friction component that is greater than the weight component alone because of the tension component perpendicular to the incline:
µ(gcos + (T/m)sin)). So
0 = g*sin - (T/m)*cos - µ*(g*cos + (T/m)*sin)
Plugging in numbers,
0 = 9.8*sin33 - (T/9.4)*cos33 - 0.374*(9.8*cos33 + (T/9.4)*sin33)
which solves to T = 20.42 N

5) I'll finally have to work with forces instead of accelerations.
For the first block:
a = g*(sin - µ*cos) - (T/m) assumes the string is now parallel to the incline
Since a = 0,
T = 9.4kg * 9.8m/s² * (sin33 - 0.374*cos33) = 21.27 N

Now for the second mass, then
a = g*(sin - µ*cos) + (T/M)
0 = 9.8*(sin33 - 0.97*cos33) + 21.27/M
I get M = 8.04kg.

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