Question 4.2a: The 5.40-kg block shown in the figure is suspended between two co

Question

Question 4.2a:
The 5.40-kg block shown in the figure is suspended between two cords of negligible mass. The cord on the left is attached to a wall and is horizontal. A second cord attached to the block has tension T = 70.0 N directed at an angle above the horizon. The block is in equilibrium.

Determine the tension Tw in the cord attaching the block to the wall and the angle .

Tw

Question 4.2b:
The two masses shown in the figure are in equilibrium and the inclined plane is frictionless.

If = 32.9° and m2 = 3.82 kg, determine the mass of m1.

_______ kg

Question 4.2c:
The three masses shown in the figure are in equilibrium and the inclined plane is frictionless.
If = 37.5°, m2 = 2.00 kg, and, m3 = 4.41 kg, determine the mass m1 and the magnitude of the tension T in the string connecting m2 and m3.

m1

Tw

Explanation / Answer

4.2a

along vertical

T*sintheta = m*g .....(1)

along horizantal

T*costheta = Tw....(2)

1^2 + 2^2

T^2 = (mg)^2 + Tw^2

70^2 = (5.4*9.81)^2 + Tw^2

Tw = 70*cos49.2 = 45.8 N <<<<<<<------answer

sintheta = (mg/)T = (5.4*9.81)/70

thet = 49.2 <<_----------------answer

++++++++++

In equilibrium

for m2

T - m2*g*sintheta = 0

T = m2*g*sintheta

for m1

m1*g - T = 0

m1*g = m2*g*sintheta

m1 = m2*sintheta

m1 = 3.82*sin32.9 = 2.07 N <<<<-----answer

________________

for m3

T23 - m3*g*sin37.5 = 0

T23 = m3*g*sin37.5

for m2

T23 + m2*g*sin37.5 - T12 = 0

m3*g*sin37.5 + m2*g*sin37.5 - T12 = 0

T12 = m3*g*sin37.5 + m2*g*sin37.5

for m1

T12 - m1*g =0

T12 = m1*g

m3*g*sin37.5 + m2*g*sin37.5 = m1*g

(4.41*9.81*sin37.5) + (2*9.81*sin37.5) = m1*9.81

m1 = 3.9 kg

T23 = m3*g*sin37.5

T23 = 4.41*9.81*sin37.5= 26.34 N

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