Jake drags a sled of a mass of 25kg across the snow. The sled is attached to a rope that he pulls wut a force of 50N at an angle of 32degrees. If there is no friction whats the normal force acting on the sled?

Guest Oct 2, 2014

The normal force is the vertical force the ground exerts on the sled to stop it sinking through the surface!  In this case the sled is partly held up by the force on the rope, so the reaction force of the ground on the sled is

normal force = mass*g - F*sin(θ)  where g is gravitational acceleration (9.8m/s2) and θ is the angle between the rope and the ground. F*sin(θ) is the vertical component of the force applied to the sled through the rope.   If we have θ = 32° then

$${\mathtt{normalforce}} = {\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{32}}^\circ\right)} \Rightarrow {\mathtt{normalforce}} = {\mathtt{218.504\: \!036\: \!788\: \!35}}$$

or normal force ≈ 218.5N

Alan  Oct 3, 2014

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