Question

A sledge has mass 15 kg. A horizontal pull of 25 N will just move the sledge when it is on a horizontal surface of compacted snow. a) Draw a diagram showing the forces acting on the sledge, modelled as a particle, when it is just on the point of sliding on this horizontal surface. b) Find a value for , the coefficient of friction between the surfaces. pls help me fast. ill mark him/ her as brainliest and other things

Answer 1

The diagram given below shows the forces acting on the sledge when it is just on the point of sliding on the horizontal surface.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\framebox(15,15){\sf{15\ kg}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(2,0){28}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(31,6.5){\sf{25\ N}}\put(7.5,0){\vector(0,-1){15}}\put(2,-19){\sf{15g\ N}}\put(7.5,15){\vector(0,1){15}}\put(6.3,31.5){\sf{R}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\end{picture}$

This diagram is called the free body diagram of the sledge, from which we can find out that,

• the 25 N force acting horizontally rightwards is the horizontal pull given to the sledge which makes it just move on the surface.

• the 15g N force acting vertically downwards is the weight of the sledge.

• the force R acting vertically upwards is the normal reaction on the sledge due to the horizontal surface.

• the force f acting horizontally leftwards is the frictional force acting on the sledge due to the horizontal surface.

The sledge experiences no vertical motion, hence the net vertical force acting on it is zero.

Thus from the diagram,

$\longrightarrow\sf{R=15g\ N}$

Taking $\sf{g=9.8\ m\,s^{-2},}$

$\longrightarrow\sf{R=15\times9.8\ N}$

$\longrightarrow\sf{R=147\ N\quad\quad\dots(1)}$

If the horizontal pull makes the sledge just move over the surface, then its magnitude should be equal to the frictional force acting on it.

This also means the net horizontal force acting on the sledge should be zero.

So from the diagram,

$\longrightarrow\sf{f=25\ N\quad\quad\dots(2)}$

Let $\mu$ be the coefficient of friction, thus,

$\longrightarrow\sf{f=\mu\,R}$

So (2) becomes,

$\longrightarrow\sf{\mu\,R=25\ N}$

From (1),

$\longrightarrow\sf{147\mu\ N=25\ N}$

$\longrightarrow\sf{\mu=\dfrac{25}{147}}$

$\longrightarrow\underline{\underline{\sf{\mu=0.17}}}$