Answer:
32.13 N
Explanation:
Given that
mass of the crate, m = 3 kg
angle of inclination, = 35°
coefficient of static friction, = 0.3
To solve this, we can assume that the minimum force is F Newton, then use the formula
mgsinA = coefficient of static friction * [F + mgcosA]
=>3 * 9.8 * sin35 = 0.3 * [F + 3 * 9.8 * cos35]
=> 29.4 * 0.5736 = 0.3 * [F + 29.4 * 0.8192]
=> 16.86 = 0.3 [F + 24.08]
=> 16.86 = 0.3F + 7.22
=> 16.86 - 7.22 = 0.3F
=> 0.3F = 9.64
=> F = 9.64/0.3
=> F = 32.13 N
Therefore, the Force that must be applied is 32.13 N
The net force that must be applied is 9.8 N.
The minimum force required is Fnet.
Fnet = -Ff + mgsinθ
But Ff = μN = μmgcosθ
Fnet = - μmgcosθ + mgsinθ
Where;
m = 3.00 kg
μ = 0.300
θ = 35.0o
Substituting values;
Fnet = mgsinθ - μmgcosθ
Fnet = (3 × 10 × sin 35.0o) - (3 × 0.300 × 10 × cos 35.0o)
= 17.2 - 7.4
Fnet = 9.8 N
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