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3 years ago

A box of mass 21 kg sits on an inclined surface with an angle of 10°. What is

Physics

2 answers:

N76 [4]3 years ago

7 0

36 N is the correct answer

alexandr1967 [171]3 years ago

4 0

The component of the weight of the box along the surface is 34.9N.

<u>Explanation:</u>

Given that

Mass of the box m=21kg

Angle of inclination θ=10°

Acceleration due to gravity g=9.8 $m/s^{2}$

Weight of an object is the force with which the earth attracts the object. Since weight is a force its unit is Newton(N). When an object is placed on a surface that has no inclination, its weight is given by the expression

W=mg

where m denotes mass of the object and g denotes the acceleration due to gravity.

When the surface has an inclination, the weight of the object can be resolved into two components that are parallel and perpendicular to the surface.

The component of weight parallel to the surface is given by

W=mgsinθ

where θ is the angle of inclination

The component of weight perpendicular to the surface is given by

W=mgcosθ

In the given problem, we have to find out the parallel component of weight

W=mgsinθ

$W=21\times 9.8 \timessin10\°\\=21\times9.8\times0.17\\=34.9N$

The component of the weight of the box along the surface is 34.9N

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Addition of vectors:

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Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

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