Question

A truck is travelling in a straight line on leveled ground and is accelerating uniformly with an acceleration of magmitude A.a rope (that can be consdered massless and inestensible)is tied to the back of the truck.The other endo of the rope is tiedto a bucket of mass M.the bucket tosses widely when the truck stays to accelerate but lateron it stays at a fixed position at a fixed distance behind the truck with the rope hanging straight at a fixed angle.
(1)find the angle theta at which the rope will settle.express your answers in terms of the given variables M,g and a
(2)the truck comes to a downhill section of the road at an angle × relative to the horizontal ,suppose that the truck continues to accelerate with an acceleration of magnitude A.once the rope again settles to a fixed angle theta relative to the truck backdoor ,what will that new angle be.
(3)what will be the new tension T in the rope be.express your answers in terms of the given variables M,g and a

Answer 1

Answer:

Part a)

The angle made by the rope with the back surface of the truck is

$\theta = tan^{-1}(\frac{a}{g})$

Part b)

Now the truck is moving downhill at an angle "alpha" then angle made by the rope is

$\phi = tan^{-1}(\frac{a - g sin\alpha}{g cos\alpha})$

Part c)

Now the new value of tension force is given as

$T = M\sqrt{a^2 + g^2}$

Explanation:

Part 1)

When truck is moving on level road then let say it makes an angle theta with the back surface of truck

So here one component of the tension force will balance the weight of the bucket and other component will accelerate in the direction of the truck

so we have

$T cos\theta = Mg$

$T sin\theta = Ma$

now we have

$tan\theta = \frac{a}{g}$

$\theta = tan^{-1}(\frac{a}{g})$

Part b)

Now the truck is moving downhill at an angle "alpha"

again we have

$T sin\phi + Mg sin\alpha = Ma$

$T cos\phi = Mg cos\alpha$

now we have

$tan\phi = \frac{Ma - Mg sin\alpha}{Mg cos\alpha}$

$\phi = tan^{-1}(\frac{a - g sin\alpha}{g cos\alpha})$

Part c)

Now the new value of tension force is given as

$(Tsin\theta)^2 + (Tcos\theta)^2 = (Mg)^2 + (Ma)^2$

$T^2 = M^2(a^2 + g^2)$

$T = M\sqrt{a^2 + g^2}$

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Topic : Tension force in the rope

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