A force is applied to an initially stationary block of mass 2.05 kg that sits on

Question

A force is applied to an initially stationary block of mass 2.05 kg that sits on a horizontal floor as shown. The 55.4 N force is applied at = 36° angle. The coefficients of friction between the floor and the block are s = 0.652 and k = 0.442. What is the acceleration of the block?

m/s2

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 7.81 kg, m2 = 7.10 kg, and = 56°. When released from rest,m1 accelerates downward at 1.262 m/s2. For this to happen, the coefficient of kinetic friction must be___________ , and to even begin sliding in the first place the coefficient of static friction must be less than_____________ .

Explanation / Answer

part 1.

normal force=weight of the block+vertical component of force

=2.05*9.8+55.4*cos(36)=64.91 N

hence static friction force=static friction coefficient*normal force

=0.652*64.91=42.321 N

horizontal component of applied force=55.4*sin(36)=32.563 N

as the driving force is less than static friction force, the block wont move

hence acceleration=0

part 2:

let coefficient of kinetic friction=k

let tension in the string be T.

writing force balance equation for m1,

m1*g-T=m1*1.262

==>T=7.81*9.8-7.81*1.262=66.682 N

normal force on m2=m2*g*sin(phi)=57.684 N

then friction force=k*nomral force=k*57.684 N

writing force balance equation for m2:

T-m2*g*cos(phi)-friction force=m2*acceleration

==>66.682-7.1*9.8*cos(56)-k*57.684=7.1*1.262

==>k=0.32614

hence coefficient of kinetic friction is 0.32614.

part b:

at the maximum value of coefficient of static friction at which the block can just begin to slide,

let tension in the string be T.

as acceleration of the sytem is 0,

writing force balance equation for m1:

T=m1*g=7.81*9.8=76.538 N

let coefficient of static friction=k
then for m2,writing force balance equation:

T=m2*g*cos(phi)+k*m2*g(sin(phi)

==>76.538=7.1*9.8*cos(56)+k*7.1*9.8*sin(56)

==>k=0.6523

hence maximum value of coefficient of static friction is 0.6523.

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