In the figure shown, the minimum force F applied perpendicular to the incline so that the block does not slide is equal to 50 X newton. Find X. (The block is always in contact with the incline)

Question

In the figure shown, the minimum force F applied perpendicular to the incline so that the block does not slide is equal to 50 X newton. Find X. (The block is always in contact with the incline) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/24e27d85cb5ce23be3438d00d834e1d4-.png" /> fs=10 g sin 37° fs=10 × 10 × (3/5)=60 N N=F-10 g cos 37° N=F-80 fs ≤ μs N 60 ≤ 0.5(F-80) 120 ≤ F-80 F ≥ 200 F min =200=50 X ∴ X=4

Q. In the figure shown, the minimum force $F$ applied perpendicular to the incline so that the block does not slide is equal to $50 X$ newton. Find $X$ . (The block is always in contact with the incline)

$f_{s} = 10 g sin 3 7^{\circ}$
$f_{s} = 10 \times 10 \times \frac{3}{5} = 60 N$
$N = F - 10 g cos 3 7^{\circ}$
$N = F - 80$
$f_{s} \leq μ_{s} N$
$60 \leq 0.5 (F - 80)$
$120 \leq F - 80$
$F \geq 200$
$F_{m i n} = 200 = 50 X$
$∴ X = 4$

Q. In the figure shown, the minimum force F applied perpendicular to the incline so that the block does not slide is equal to 50X newton. Find X. (The block is always in contact with the incline)

fs​=10gsin37∘ fs​=10×10×53​=60N N=F−10gcos37∘ N=F−80 fs​≤μs​N 60≤0.5(F−80) 120≤F−80 F≥200 Fmin​=200=50X ∴X=4

Q. In the figure shown, the minimum force $F$ applied perpendicular to the incline so that the block does not slide is equal to $50 X$ newton. Find $X$ . (The block is always in contact with the incline)

$f_{s} = 10 g sin 3 7^{\circ}$
$f_{s} = 10 \times 10 \times \frac{3}{5} = 60 N$
$N = F - 10 g cos 3 7^{\circ}$
$N = F - 80$
$f_{s} \leq μ_{s} N$
$60 \leq 0.5 (F - 80)$
$120 \leq F - 80$
$F \geq 200$
$F_{m i n} = 200 = 50 X$
$∴ X = 4$