# Find the resultant force in a motion problem with a loop

In the approximate diagram, the block with mass $$m$$ starts without movement in the position $$P$$ in the top of a hill with height $$5R$$. After that, the block falls down through the hill, and reaches a loop until it arrives at the position $$Q$$ with a height of $$R$$. If the loop is a circle of radius $$R$$ and there is no friction, what is the value of the resultant force in the point $$Q$$?

My try

I tried to use the law of conservation of energy in the points $$P$$ and $$Q$$, with this i get the speed, but i don't know how to continue in this problem.

Any hints?

• The resultant force would be the vector sum of centripetal force, and the force due to gravity – Eagle May 3 '19 at 23:05
• @Eagle the centripetal force is in function of the mass? – Rodrigo Pizarro May 4 '19 at 0:01
• The centripetal force is $mv^2/R$. Find v by energy conservation. – Tojrah May 4 '19 at 1:59

Hint: the net force will be the resultant of the normal force which is horizontal (equal to centripetal force) at A and obviously the weight(=mg)which acts vertically downwards.

• So the sum of both gives the answer? – Rodrigo Pizarro May 4 '19 at 4:09
• The vector sum, not simply adding as scalars! The vector sum of two perpendicular vectors $\vec a$ and $\vec b$ has magnitude $\sqrt{a^2+b^2}$ – Tojrah May 4 '19 at 4:12

My guess is that as it's a slope we''ll divide mg into components .Mg sintheta will provide linear acceleration whereas mgcostheta will provide centripetal acceleration. Multiply by mass and u get the force .The resultant force would be the vector sum of centripetal force, and the force due to gravity.(As said in comment )