Answer:
E.71 N
Explanation:
Given Data
Mass=12 kg
Angle=30°
static friction=0.44 N
To Find
Force=?
Solution
Sum of the forces in the direction of motion is given as
= 0 = F×cos(30) - friction
Sum of the forces in the vertical direction (up is positive) is given as
= 0 = N - mg - F×sin(30)
N = m×g + F×sin(30)
From the definition of friction
friction = u×N
friction = u×(m×g+F×sin(30))
Plug back into the top
0 = F*cos(30) - (u*m*g+u*F*sin(30))
Solve for F
u×m×g = F×(cos(30) - u×sin(30))
F = (u×m×g) / (cos(30) - u×sin(30))
Put the number we get
F = 71 N
So Option E is correct
"71 N" would be the minimum magnitude force he needs.
According to the question,
Mass,
Coefficient of static friction,
Angle,
We know,
→ [tex]\Sigma F = F_{app} -f[/tex]
[tex]0 = F Cos \Theta-f[/tex]
Force along vertical,
→ [tex]N= mg +F Sin \Theta[/tex]
or,
→ [tex]F Cos \Theta = f = \mu (mg+F Sin \Theta)[/tex]
→ [tex]F Cos \Theta- \mu F Sin \Theta = \mu mg[/tex]
[tex]F = \frac{\mu mg}{Cos \Theta - \mu Sin \Theta}[/tex]
By substituting the values, we get
[tex]= \frac{0.4\times 12\times 9.8}{Cos 30^{\circ} -(0.4) Sin 30^{\circ}}[/tex]
[tex]= 70.62 \ N[/tex]
or,
[tex]= 71 \ N[/tex]
Thus the above answer i.e., "Option E" is right.
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