(c) A ladder with length L and mass m rests against a wall. Its upper end is a distance h above the ground (Figure 13.3). The center of gravity of the ladder is one-third of the way up the ladder. A fire fighter with mass M climbs halfway up the ladder. Assume that the wall, but not the ground, is frictionless. Write the sum of the forces in (i) x-direction and (ii) y-direction and (iii) the expression for the sum of the torque about the point where the ladder makes contact with the ground(O).
(a) A 5.00 kg block is moving at vo= 6.00m/s along a frictionless, horizontal surface towards a spring with a force constant k = 500 N/m that is attached to a wall (Figure 13.5). The spring has negligible mass. Find(i) the maximum distance the spring will be compressed and (ii) the spring is compressed by no more than 0.150 m, what should be the maximum value of vo? (4 marks)
(b) A steel railroad track is 1000 m long. When its temperature is raised from -30 oto + 30 othermal expansion causes the length of the track to increase by .01%. What compressive force per unit area would be required to keep the track from expanding? (Hint: Calculate the force that would be required to compress it back to its original length. Young's modulus for steel is 2 x 1011N/m2.
(c) A hydraulic press contains 0.25m3(250 L) of oil. Find the decrease in volume of the oil when it is subjected to a pressure increase, ∆p = 1.6 x 107 Pa. The bulk modulus , B of oil is 5.0 x 109 Pa and its compressibility K = 1/B = 2.0 x 10-10 Pa-1.
To determine the sum of the forces in the x-direction, y-direction, and the torque about the point of contact with the ground (O), we can break down the forces acting on the ladder.
(i) Sum of forces in the x-direction:
Since the wall is frictionless, there are no horizontal forces acting on the ladder. Therefore, the sum of the forces in the x-direction is zero.
(ii) Sum of forces in the y-direction:
Considering the forces acting on the ladder, we have:
- The weight of the ladder (mg) acting vertically downward.
- The weight of the fire fighter (Mg) acting vertically downward.
- A normal force exerted by the ground in the upward direction.
The sum of the forces in the y-direction is given by:
ΣFy = N - mg - Mg
(iii) Sum of torque about the point of contact (O):
To find the torque, we need to consider the distances from the point of contact to the forces acting on the ladder. Let's assume the length of the ladder is L, and the distance from the center of gravity to the point of contact is d.
The torques can be calculated using the following formula:
τ = F * r * sin(θ)
Considering the forces and torques acting on the ladder, we have:
- The weight of the ladder (mg) acting at a distance of (L/2) from point O.
- The weight of the fire fighter (Mg) acting at a distance of (L/4) from point O.
- The normal force (N) acting at a distance of 0 from point O (since it is at the same point).
The sum of the torques about point O is given by:
ΣτO = (mg * (L/2) * sin(90°)) + (Mg * (L/4) * sin(90°))
Note: The sin(90°) term simplifies to 1, as sin(90°) = 1.
This is the expression for the sum of the torque about the point where the ladder makes contact with the ground (O).