Complete question is;
Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?
Answer:
F = 501.5 N
Explanation:
We are given;
Mass of wooden crate; m = 75 kg
Angle of ramp; θ = 11°
Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;
F = mgsin θ + μmg cos θ
From online values, coefficient of friction between wooden surfaces is μ = 0.5
Thus;
F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)
F = 501.5 N
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Answer:
the rotational inertia of the cylinder = 4.85 kgm²
the mass moved 7.942 m/s
Explanation:
Formula for calculating Inertia can be expressed as:
For calculating the rotational inertia of the cylinder ; we have;
I ≅ 4.85 kgm²
mg - T ma and RT = I ∝
T = 
a = 4.1713 m/s²
Using the equation of motion
The resultant of the given forces is; 6√2 N
<h3>How to find the resultant of forces</h3>
We are given the forces as;
10 N along the x-axis which is +10 N in the x-direction
6 N along the y-axis which is +6N in the y-direction
4 N along the negative x-axis which is -4N
Thus;
Resultant force in the x-direction is; 10 - 4 = 6N
Resultant force in the y-direction is; 6N
Thus;
Total resultant force = √(6² + 6²)
Total resultant force = 6√2 N
Read more about finding resultant of a force at; brainly.com/question/14626208
