1 . A 1240 N uniform beam is attached to a vertical wall at one end and is suppo

Question

1. A 1240 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1990 N crate hangs from the far end of the beam.
a) Calculate the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam if the angle between the cable and horizontal is = 36°.
N

2. A mass m1 = 16.5 kg is connected by a light string that passes over a pulley of mass M = 15.0 kg to a mass m2 = 5.00 kg sliding on an inclined surface that makes an angle of 26° with the horizontal (see figure). The coefficient of kinetic friction between the mass m2 and the surface is 0.20. There is no slippage between the string and the pulley. What is the magnitude of the acceleration of the system of masses? (The moment of inertia of the pulley is ½Mr².)
m/s²

3. A solid sphere starts from rest at the upper end of the track, as shown in figure below, and rolls without slipping until it rolls off the right-hand end. If H= 12.0 m and h = 6.00 m and the track is horizontal at the right-hand end, how far horizontally from point A does the sphere land on the floor?
m

Explanation / Answer

1)

In equilibrium net torque = 0

T*costheta*L*sin30 + T*sintheta*L*cos30 - Wbeam*L/2*cos30 - Wcrate*L*cos30 = 0

T*sin(theta+30) - Wbeam/2*cos30 - Wcrate*cos30 = 0

T*sin(36+30) - 1240/2*cos30 - 1990*cos30 = 0

T = 2474.23 N

along horizontal

Fx - Tx = 0

Fx = Tx = 2474.23*cos30 = 2142.75 N <<<<<<-----ANSWER

==========================================

from newtons second law Fnet = m*a

for m1

m1*g - T1 = m1*a

T1 = m1*g - m1*a

for m2

T2 - m2*g*sintheta - uk*m2*g*costheta = m2*a

T2 = m2*g*sintheta + uk*m2*g*costheta + m2*a

for pulley

(T1 - T2)*R = I*alpha

angular acceleration = alpha = a/R

moment of inertia = I = (1/2)*M*R^2

(T1 - T2)*R = (1/2)*M*R^2*a/R

T1 - T2 = (1/2)*M*a

m1*g - m1*a - (m2*g*sintheta + uk*m2*g*costheta + m2*a) = (1/2)*M*a

m1*g - m1*a - m2*g*sintheta - uk*m2*g*costheta - m2*a = (1/2)*M*a

a = g*(m1 - m2*sintheta - uk*m2*costheta)/(m1 + m2 + M/2 )

a = 9.8*(16.5 - (5*sin26) - (0.2*5*cos26))/(16.5 + 5 + 15/2)

acceleration a = 4.53 m/s^2

======================================

potential energy at H = Kinetic energy at h

m*g*(H - h) = (1/2)*I*w^2 + (1/2)*m*v^2

I = (2/5)*M*R^2

w = v/R

g*(H-h) = (1/5)*v^2 + (1/2)*v^2

g*(H-h) = (7/10)*v^2

v = sqrt(10*g*(H-h))/7)

v = sqrt(10*9.8*(12-6)/7)

v = 9.16 m/s

after rolling off from h

along vertical

-h = -(1/2)*g*t^2

t = sqrt(2h/g)

along horizontal

x = v*t = v*sqrt(2h/g)

x = 9.16*sqrt(2*6/9.8)

x = 10.14 m <<<<<---------ANSWER

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