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miv72 [106K]
3 years ago
14

You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what

magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

<em>345.60N</em>

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

<em>Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N</em>

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
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Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

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F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

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