The maximum value the string tension can have before the can slip is 17.80 N
Let consider the mass of the can by making an assumption since the mass is not given.
- assuming the mass of the can = 3.0 kg
- assuming the angle formed = 40°
- the coefficient of static friction [tex]\mathbf{\mu_s= 0.76}[/tex]
By resolving the components of the angle;
[tex]\sum F_y = 0[/tex]
∴
- N + Tsin θ = mg
- N = mg - Tsinθ ------- (1)
The static friction [tex]\mathbf{F_s}[/tex] acting in the opposite direction is:
[tex]\mathbf{F_s = \mu_sN}[/tex]
[tex]\mathbf{F_s = \mu_s(mg - Tsin \theta)}[/tex]
Thus, the maximum value of the tension in the string before it slips can be expressed as:
[tex]\mathbf{Tcos \theta =F_s}[/tex]
[tex]\mathbf{T cos \theta= \mu_s(mg - Tsin \theta)}[/tex]
[tex]\mathbf{T cos \theta+ \mu_s Tsin \theta = \mu_s mg}[/tex]
[tex]\mathbf{T( cos \theta+ \mu_s sin \theta )= \mu_s mg}[/tex]
[tex]\mathbf{T=\dfrac{ \mu_s mg}{( cos \theta+ \mu_s sin \theta )}}[/tex]
[tex]\mathbf{T=\dfrac{(0.76 ) \times 3 \times 9.8}{( cos 40)+ ( 0.76 \times sin 40) )}}[/tex]
[tex]\mathbf{T=\dfrac{22.344}{0.7660+0.4885 }}[/tex]
[tex]\mathbf{T=\dfrac{22.344}{1.2545}}[/tex]
T = 17.80 N
Therefore, the maximum value the string tension can have before the can slip is 17.80 N
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