You push against a steamer trunk at an angle of 33 degrees with the horizontal . The trunk is on a flat floor and the coefficient of static friction between the trunk and floor is 0.55. The mass of the trunk is 43 kg. What is the minimum magnitude of force that will allow you to move the trunk?

Question

anonymous · Answer

Are not you given the force acting on that angle?

anonymous · Answer

no @Hoslos

anonymous · Answer

Could you tell me what the answer is?

anonymous · Answer

the answers are:

        300 N
	280 N
	350 N
	220 N

anonymous · Answer

Ohh, I got it now. I pictured the situation wrongly. 
So first things first is to find normal contact force (N), which in this case because the floor is flat is the same as weight or mg or 43*9.81= 421.83N
Ff(frictional force)=€N, where € is coefficient of friction.=0.55*421.83=232N.
So the minimum force has to be exactly as the frictional one, but it is said that the force that you apply is at 33o to the horizontal, therefore the equation is Fcos33=232, then F= 276.6 N, reasonably your 280 answer.

anonymous · Answer

thank you

anonymous · Answer

What about:

You push against a steamer trunk with a force of 750 N at an angle of 25 degrees with the horizontal. The trunk is on a flat floor and the coefficient of static friction between the trunk and floor is 0.77. What is the most massive trunk you will be able to move?

anonymous · Answer

Again, you first find N: mg or 9.81m Newtons.
Ff=€N or 0.77*9.81m=7.55m Newtons
Again Ff must be equal to driving force or 750cos25=7.55m and m=90kg