you want to nail a 1.6 kg board onto the wall of a barn. to position the board before nailing, you push it against the wall with a horizontal force to keep it from sliding to the ground. (a) if the coefficient of static friction between the board and wall is .79, what is the least froce you can apply and still hold the board in place? (b) what hppens to the force of static friction if you push against the wall with a force greater that that found in part a?

Question

Anonymous · Accepted Answer

OK so it's hard to draw a free-body diagram on a computer but it would be a the board with its weight(mg) pulling it down, the frictional force holding it up, the Normal force being one horizontal component, and the Force you're applying being the opposite horizontal component. Since the object is in equilibrium you can use the 2 equations
F(static)=weight(mg) and Normal=Force applied
and then the frictional force formula
F(static)=constant*Normal
so substituting equations,
F(static)=constant*Force applied
weight=constant*Force applied
Force applied=mg/constant
F applied=(1.6)(9.8)/.79=19.8N

and as you can see from the previous work, increasing the force applied increases the normal force which then increases the frictional force

Anonymous · Answer

You don't divide by .79, you multiply. The correct answer is 12.4N

anonymous · Answer

Just tried it on my hw, and you do divide. Multiplying gave me the wrong answer.