Question

find the force which a) acting horizontally b) acting at an angle of 45 with the horizontal will just pull a body weighing 5 kg along a rough horizontal surface limiting angle of friction between the body and the plane is 30

Please solve this on the paper and don't post irrelevant answer guys be humans ​

Answer 1

Proper Question: Find the force which is

a) acting horizontally

b) acting at an angle of 45° with the horizontal

will just pull a body weighing 5 kg along a rough horizontal surface limiting the angle of friction between the body and the plane is 30°.

Answers:

(a) In case the force acting is in the horizontal direction, the force required would be equal to 50/√3 Newton.

(b) In case the force acting is acting at an angle of 45° with the horizontal, the force required would be equal to 50√2 /(√3+1) Newton.

Given:

Mass of the body (m) = 5 kg

Limiting angle of friction between the body and the plane (α) = 30°

To Find:

The force that will just pull a body weighing 5 kg along a rough horizontal surface. In case the force acting is:

a) acting horizontally

b) acting at an angle of 45° with the horizontal.

Solution:

∵ The limiting angle of friction between the body and the plane (α) is 30°.

∴ The coefficient of friction (μ) would be equal to tan(α).

∴ The coefficient of friction (μ) = tan(30°) = 1/√3

(a) In case the force acting is in the horizontal direction:

∴ The static friction (f) = μN = μ (mg) = (1/√3)(5)(10) = 50/√3 N

→ For the force (F) to just pull the body: 'F' must be equal to the static friction (f):

∴ F = f = 50/√3 N

Therefore the force required would be equal to 50/√3 Newton.

(b) In case the force acting is acting at an angle of 45° with the horizontal:

∴ The static friction (f) = μN = μ (mg) = (1/√3)(5)(10) = 50/√3 N

→ For the force (F) to just pull the body: The horizontal component of the force 'F' must be equal to the static friction (f):

                         $\therefore Fcos45^{0} = f=\mu N \\\\\therefore Fcos45^{0} =\frac{1}{\sqrt{3} } (50-Fsin45^{0} ) \\\\\therefore F(\frac{1}{\sqrt{2} } ) = \frac{50}{\sqrt{3} } - F(\frac{1}{\sqrt{2} \times\sqrt{3} } ) \\\\\therefore F(\frac{(\sqrt{3}+1) }{\sqrt{6} } ) = 50(\frac{1}{\sqrt{3} }) \\\\ \therefore F=(\frac{50\sqrt{2}}{\sqrt{3}+1} ) \:Newton$

Therefore the force required would be equal to 50√2 /(√3+1) Newton.

(a) In case the force acting is in the horizontal direction, the force required would be equal to 50/√3 Newton. (b) In case the force acting is acting at an angle of 45° with the horizontal, the force required would be equal to 50√2 /(√3+1) Newton.

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